Solve 5(4,25-y)^2=320y

$$5 ( 4,25 - y ) ^ { 2 } = 320 y$$

Solve for y

$$y=72,25$$

$$y=0,25$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4,25-y\right)^{2}.

$$5\left(18,0625-8,5y+y^{2}\right)=320y$$

Use the distributive property to multiply 5 by 18,0625-8,5y+y^{2}.

$$90,3125-42,5y+5y^{2}=320y$$

Subtract 320y from both sides.

$$90,3125-42,5y+5y^{2}-320y=0$$

Combine -42,5y and -320y to get -362,5y.

$$90,3125-362,5y+5y^{2}=0$$

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.

$$5y^{2}-362,5y+90,3125=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -362,5 for b, and 90,3125 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-362,5\right)±\sqrt{\left(-362,5\right)^{2}-4\times 5\times 90,3125}}{2\times 5}$$

Square -362,5 by squaring both the numerator and the denominator of the fraction.

$$y=\frac{-\left(-362,5\right)±\sqrt{131406,25-4\times 5\times 90,3125}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-362,5\right)±\sqrt{131406,25-20\times 90,3125}}{2\times 5}$$

Multiply -20 times 90,3125.

$$y=\frac{-\left(-362,5\right)±\sqrt{\frac{525625-7225}{4}}}{2\times 5}$$

Add 131406,25 to -1806,25 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-362,5\right)±\sqrt{129600}}{2\times 5}$$

Take the square root of 129600.

$$y=\frac{-\left(-362,5\right)±360}{2\times 5}$$

The opposite of -362,5 is 362,5.

$$y=\frac{362,5±360}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{362,5±360}{10}$$

Now solve the equation y=\frac{362,5±360}{10} when ± is plus. Add 362,5 to 360.

$$y=\frac{722,5}{10}$$

Divide 722,5 by 10.

$$y=72,25$$

Now solve the equation y=\frac{362,5±360}{10} when ± is minus. Subtract 360 from 362,5.

$$y=\frac{2,5}{10}$$

Divide 2,5 by 10.

$$y=0,25$$

The equation is now solved.

$$y=72,25 y=0,25$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4,25-y\right)^{2}.

$$5\left(18,0625-8,5y+y^{2}\right)=320y$$

Use the distributive property to multiply 5 by 18,0625-8,5y+y^{2}.

$$90,3125-42,5y+5y^{2}=320y$$

Subtract 320y from both sides.

$$90,3125-42,5y+5y^{2}-320y=0$$

Combine -42,5y and -320y to get -362,5y.

$$90,3125-362,5y+5y^{2}=0$$

Subtract 90,3125 from both sides. Anything subtracted from zero gives its negation.

$$-362,5y+5y^{2}=-90,3125$$

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.

$$5y^{2}-362,5y=-90,3125$$

Divide both sides by 5.

$$\frac{5y^{2}-362,5y}{5}=-\frac{90,3125}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$y^{2}+\left(-\frac{362,5}{5}\right)y=-\frac{90,3125}{5}$$

Divide -362,5 by 5.

$$y^{2}-72,5y=-\frac{90,3125}{5}$$

Divide -90,3125 by 5.

$$y^{2}-72,5y=-18,0625$$

Divide -72,5, the coefficient of the x term, by 2 to get -36,25. Then add the square of -36,25 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-72,5y+\left(-36,25\right)^{2}=-18,0625+\left(-36,25\right)^{2}$$

Square -36,25 by squaring both the numerator and the denominator of the fraction.

$$y^{2}-72,5y+1314,0625=\frac{-289+21025}{16}$$

Add -18,0625 to 1314,0625 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-72,5y+1314,0625=1296$$

Factor y^{2}-72,5y+1314,0625. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-36,25\right)^{2}=1296$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-36,25\right)^{2}}=\sqrt{1296}$$

Simplify.

$$y-36,25=36 y-36,25=-36$$

Add 36,25 to both sides of the equation.

$$y=72,25 y=0,25$$