Solve 0.5*10^-3=1/2*350*3.83*10^-7*25*(3-y-0.7)^2

$$0.5 \times 10 ^ { - 3 } = \frac { 1 } { 2 } \times 350 \times 3.83 \times 10 ^ { - 7 } \times 25 \times ( 3 - y - 0.7 ) ^ { 2 }$$

Solve for y

$$y=-\frac{20\sqrt{5362}}{2681}+2.3\approx 1.753743539$$

$$y=\frac{20\sqrt{5362}}{2681}+2.3\approx 2.846256461$$

Steps Using the Quadratic Formula

Cancel out 0.5 on both sides.

$$10^{-3}=350\times 3.83\times 10^{-7}\times 25\left(3-y-0.7\right)^{2}$$

Calculate 10 to the power of -3 and get \frac{1}{1000}.

$$\frac{1}{1000}=350\times 3.83\times 10^{-7}\times 25\left(3-y-0.7\right)^{2}$$

Multiply 350 and 3.83 to get 1340.5.

$$\frac{1}{1000}=1340.5\times 10^{-7}\times 25\left(3-y-0.7\right)^{2}$$

Calculate 10 to the power of -7 and get \frac{1}{10000000}.

$$\frac{1}{1000}=1340.5\times \frac{1}{10000000}\times 25\left(3-y-0.7\right)^{2}$$

Multiply 1340.5 and \frac{1}{10000000} to get \frac{2681}{20000000}.

$$\frac{1}{1000}=\frac{2681}{20000000}\times 25\left(3-y-0.7\right)^{2}$$

Multiply \frac{2681}{20000000} and 25 to get \frac{2681}{800000}.

$$\frac{1}{1000}=\frac{2681}{800000}\left(3-y-0.7\right)^{2}$$

Subtract 0.7 from 3 to get 2.3.

$$\frac{1}{1000}=\frac{2681}{800000}\left(2.3-y\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2.3-y\right)^{2}.

$$\frac{1}{1000}=\frac{2681}{800000}\left(5.29-4.6y+y^{2}\right)$$

Use the distributive property to multiply \frac{2681}{800000} by 5.29-4.6y+y^{2}.

$$\frac{1}{1000}=\frac{1418249}{80000000}-\frac{61663}{4000000}y+\frac{2681}{800000}y^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1418249}{80000000}-\frac{61663}{4000000}y+\frac{2681}{800000}y^{2}=\frac{1}{1000}$$

Subtract \frac{1}{1000} from both sides.

$$\frac{1418249}{80000000}-\frac{61663}{4000000}y+\frac{2681}{800000}y^{2}-\frac{1}{1000}=0$$

Subtract \frac{1}{1000} from \frac{1418249}{80000000} to get \frac{1338249}{80000000}.

$$\frac{1338249}{80000000}-\frac{61663}{4000000}y+\frac{2681}{800000}y^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{2681}{800000}y^{2}-\frac{61663}{4000000}y+\frac{1338249}{80000000}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{2681}{800000} for a, -\frac{61663}{4000000} for b, and \frac{1338249}{80000000} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{61663}{4000000}\right)±\sqrt{\left(-\frac{61663}{4000000}\right)^{2}-4\times \frac{2681}{800000}\times \frac{1338249}{80000000}}}{2\times \frac{2681}{800000}}$$

Square -\frac{61663}{4000000} by squaring both the numerator and the denominator of the fraction.

$$y=\frac{-\left(-\frac{61663}{4000000}\right)±\sqrt{\frac{3802325569}{16000000000000}-4\times \frac{2681}{800000}\times \frac{1338249}{80000000}}}{2\times \frac{2681}{800000}}$$

Multiply -4 times \frac{2681}{800000}.

$$y=\frac{-\left(-\frac{61663}{4000000}\right)±\sqrt{\frac{3802325569}{16000000000000}-\frac{2681}{200000}\times \frac{1338249}{80000000}}}{2\times \frac{2681}{800000}}$$

Multiply -\frac{2681}{200000} times \frac{1338249}{80000000} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{61663}{4000000}\right)±\sqrt{\frac{3802325569-3587845569}{16000000000000}}}{2\times \frac{2681}{800000}}$$

Add \frac{3802325569}{16000000000000} to -\frac{3587845569}{16000000000000} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{61663}{4000000}\right)±\sqrt{\frac{2681}{200000000}}}{2\times \frac{2681}{800000}}$$

Take the square root of \frac{2681}{200000000}.

$$y=\frac{-\left(-\frac{61663}{4000000}\right)±\frac{\sqrt{5362}}{20000}}{2\times \frac{2681}{800000}}$$

The opposite of -\frac{61663}{4000000} is \frac{61663}{4000000}.

$$y=\frac{\frac{61663}{4000000}±\frac{\sqrt{5362}}{20000}}{2\times \frac{2681}{800000}}$$

Multiply 2 times \frac{2681}{800000}.

$$y=\frac{\frac{61663}{4000000}±\frac{\sqrt{5362}}{20000}}{\frac{2681}{400000}}$$

Now solve the equation y=\frac{\frac{61663}{4000000}±\frac{\sqrt{5362}}{20000}}{\frac{2681}{400000}} when ± is plus. Add \frac{61663}{4000000} to \frac{\sqrt{5362}}{20000}.

$$y=\frac{\frac{\sqrt{5362}}{20000}+\frac{61663}{4000000}}{\frac{2681}{400000}}$$

Divide \frac{61663}{4000000}+\frac{\sqrt{5362}}{20000} by \frac{2681}{400000} by multiplying \frac{61663}{4000000}+\frac{\sqrt{5362}}{20000} by the reciprocal of \frac{2681}{400000}.

$$y=\frac{20\sqrt{5362}}{2681}+\frac{23}{10}$$

Now solve the equation y=\frac{\frac{61663}{4000000}±\frac{\sqrt{5362}}{20000}}{\frac{2681}{400000}} when ± is minus. Subtract \frac{\sqrt{5362}}{20000} from \frac{61663}{4000000}.

$$y=\frac{-\frac{\sqrt{5362}}{20000}+\frac{61663}{4000000}}{\frac{2681}{400000}}$$

Divide \frac{61663}{4000000}-\frac{\sqrt{5362}}{20000} by \frac{2681}{400000} by multiplying \frac{61663}{4000000}-\frac{\sqrt{5362}}{20000} by the reciprocal of \frac{2681}{400000}.

$$y=-\frac{20\sqrt{5362}}{2681}+\frac{23}{10}$$

The equation is now solved.

$$y=\frac{20\sqrt{5362}}{2681}+\frac{23}{10} y=-\frac{20\sqrt{5362}}{2681}+\frac{23}{10}$$

Steps for Completing the Square

Cancel out 0.5 on both sides.

$$10^{-3}=350\times 3.83\times 10^{-7}\times 25\left(3-y-0.7\right)^{2}$$

Calculate 10 to the power of -3 and get \frac{1}{1000}.

$$\frac{1}{1000}=350\times 3.83\times 10^{-7}\times 25\left(3-y-0.7\right)^{2}$$

Multiply 350 and 3.83 to get 1340.5.

$$\frac{1}{1000}=1340.5\times 10^{-7}\times 25\left(3-y-0.7\right)^{2}$$

Calculate 10 to the power of -7 and get \frac{1}{10000000}.

$$\frac{1}{1000}=1340.5\times \frac{1}{10000000}\times 25\left(3-y-0.7\right)^{2}$$

Multiply 1340.5 and \frac{1}{10000000} to get \frac{2681}{20000000}.

$$\frac{1}{1000}=\frac{2681}{20000000}\times 25\left(3-y-0.7\right)^{2}$$

Multiply \frac{2681}{20000000} and 25 to get \frac{2681}{800000}.

$$\frac{1}{1000}=\frac{2681}{800000}\left(3-y-0.7\right)^{2}$$

Subtract 0.7 from 3 to get 2.3.

$$\frac{1}{1000}=\frac{2681}{800000}\left(2.3-y\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2.3-y\right)^{2}.

$$\frac{1}{1000}=\frac{2681}{800000}\left(5.29-4.6y+y^{2}\right)$$

Use the distributive property to multiply \frac{2681}{800000} by 5.29-4.6y+y^{2}.

$$\frac{1}{1000}=\frac{1418249}{80000000}-\frac{61663}{4000000}y+\frac{2681}{800000}y^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1418249}{80000000}-\frac{61663}{4000000}y+\frac{2681}{800000}y^{2}=\frac{1}{1000}$$

Subtract \frac{1418249}{80000000} from both sides.

$$-\frac{61663}{4000000}y+\frac{2681}{800000}y^{2}=\frac{1}{1000}-\frac{1418249}{80000000}$$

Subtract \frac{1418249}{80000000} from \frac{1}{1000} to get -\frac{1338249}{80000000}.

$$-\frac{61663}{4000000}y+\frac{2681}{800000}y^{2}=-\frac{1338249}{80000000}$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{2681}{800000}y^{2}-\frac{61663}{4000000}y=-\frac{1338249}{80000000}$$

Divide both sides of the equation by \frac{2681}{800000}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{2681}{800000}y^{2}-\frac{61663}{4000000}y}{\frac{2681}{800000}}=-\frac{\frac{1338249}{80000000}}{\frac{2681}{800000}}$$

Dividing by \frac{2681}{800000} undoes the multiplication by \frac{2681}{800000}.

$$y^{2}+\left(-\frac{\frac{61663}{4000000}}{\frac{2681}{800000}}\right)y=-\frac{\frac{1338249}{80000000}}{\frac{2681}{800000}}$$

Divide -\frac{61663}{4000000} by \frac{2681}{800000} by multiplying -\frac{61663}{4000000} by the reciprocal of \frac{2681}{800000}.

$$y^{2}-\frac{23}{5}y=-\frac{\frac{1338249}{80000000}}{\frac{2681}{800000}}$$

Divide -\frac{1338249}{80000000} by \frac{2681}{800000} by multiplying -\frac{1338249}{80000000} by the reciprocal of \frac{2681}{800000}.

$$y^{2}-\frac{23}{5}y=-\frac{1338249}{268100}$$

Divide -\frac{23}{5}, the coefficient of the x term, by 2 to get -\frac{23}{10}. Then add the square of -\frac{23}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{23}{5}y+\left(-\frac{23}{10}\right)^{2}=-\frac{1338249}{268100}+\left(-\frac{23}{10}\right)^{2}$$

Square -\frac{23}{10} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{23}{5}y+\frac{529}{100}=-\frac{1338249}{268100}+\frac{529}{100}$$

Add -\frac{1338249}{268100} to \frac{529}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{23}{5}y+\frac{529}{100}=\frac{800}{2681}$$

Factor y^{2}-\frac{23}{5}y+\frac{529}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{23}{10}\right)^{2}=\frac{800}{2681}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{23}{10}\right)^{2}}=\sqrt{\frac{800}{2681}}$$

Simplify.

$$y-\frac{23}{10}=\frac{20\sqrt{5362}}{2681} y-\frac{23}{10}=-\frac{20\sqrt{5362}}{2681}$$

Add \frac{23}{10} to both sides of the equation.

$$y=\frac{20\sqrt{5362}}{2681}+\frac{23}{10} y=-\frac{20\sqrt{5362}}{2681}+\frac{23}{10}$$