Solve 30t=2.25(t+10)^2

$$30 t = 2.25 ( t + 10 ) ^ { 2 }$$

Solve for t

$$t=\frac{-10+i\times 20\sqrt{2}}{3}\approx -3.333333333+9.428090416i$$

$$t=\frac{-i\times 20\sqrt{2}-10}{3}\approx -3.333333333-9.428090416i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+10\right)^{2}.

$$30t=2.25\left(t^{2}+20t+100\right)$$

Use the distributive property to multiply 2.25 by t^{2}+20t+100.

$$30t=2.25t^{2}+45t+225$$

Subtract 2.25t^{2} from both sides.

$$30t-2.25t^{2}=45t+225$$

Subtract 45t from both sides.

$$30t-2.25t^{2}-45t=225$$

Combine 30t and -45t to get -15t.

$$-15t-2.25t^{2}=225$$

Subtract 225 from both sides.

$$-15t-2.25t^{2}-225=0$$

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.

$$-2.25t^{2}-15t-225=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2.25 for a, -15 for b, and -225 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-2.25\right)\left(-225\right)}}{2\left(-2.25\right)}$$

Square -15.

$$t=\frac{-\left(-15\right)±\sqrt{225-4\left(-2.25\right)\left(-225\right)}}{2\left(-2.25\right)}$$

Multiply -4 times -2.25.

$$t=\frac{-\left(-15\right)±\sqrt{225+9\left(-225\right)}}{2\left(-2.25\right)}$$

Multiply 9 times -225.

$$t=\frac{-\left(-15\right)±\sqrt{225-2025}}{2\left(-2.25\right)}$$

Add 225 to -2025.

$$t=\frac{-\left(-15\right)±\sqrt{-1800}}{2\left(-2.25\right)}$$

Take the square root of -1800.

$$t=\frac{-\left(-15\right)±30\sqrt{2}i}{2\left(-2.25\right)}$$

The opposite of -15 is 15.

$$t=\frac{15±30\sqrt{2}i}{2\left(-2.25\right)}$$

Multiply 2 times -2.25.

$$t=\frac{15±30\sqrt{2}i}{-4.5}$$

Now solve the equation t=\frac{15±30\sqrt{2}i}{-4.5} when ± is plus. Add 15 to 30i\sqrt{2}.

$$t=\frac{15+30\sqrt{2}i}{-4.5}$$

Divide 15+30i\sqrt{2} by -4.5 by multiplying 15+30i\sqrt{2} by the reciprocal of -4.5.

$$t=\frac{-20\sqrt{2}i-10}{3}$$

Now solve the equation t=\frac{15±30\sqrt{2}i}{-4.5} when ± is minus. Subtract 30i\sqrt{2} from 15.

$$t=\frac{-30\sqrt{2}i+15}{-4.5}$$

Divide 15-30i\sqrt{2} by -4.5 by multiplying 15-30i\sqrt{2} by the reciprocal of -4.5.

$$t=\frac{-10+20\sqrt{2}i}{3}$$

The equation is now solved.

$$t=\frac{-20\sqrt{2}i-10}{3} t=\frac{-10+20\sqrt{2}i}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+10\right)^{2}.

$$30t=2.25\left(t^{2}+20t+100\right)$$

Use the distributive property to multiply 2.25 by t^{2}+20t+100.

$$30t=2.25t^{2}+45t+225$$

Subtract 2.25t^{2} from both sides.

$$30t-2.25t^{2}=45t+225$$

Subtract 45t from both sides.

$$30t-2.25t^{2}-45t=225$$

Combine 30t and -45t to get -15t.

$$-15t-2.25t^{2}=225$$

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.

$$-2.25t^{2}-15t=225$$

Divide both sides of the equation by -2.25, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-2.25t^{2}-15t}{-2.25}=\frac{225}{-2.25}$$

Dividing by -2.25 undoes the multiplication by -2.25.

$$t^{2}+\left(-\frac{15}{-2.25}\right)t=\frac{225}{-2.25}$$

Divide -15 by -2.25 by multiplying -15 by the reciprocal of -2.25.

$$t^{2}+\frac{20}{3}t=\frac{225}{-2.25}$$

Divide 225 by -2.25 by multiplying 225 by the reciprocal of -2.25.

$$t^{2}+\frac{20}{3}t=-100$$

Divide \frac{20}{3}, the coefficient of the x term, by 2 to get \frac{10}{3}. Then add the square of \frac{10}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{20}{3}t+\frac{10}{3}^{2}=-100+\frac{10}{3}^{2}$$

Square \frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{20}{3}t+\frac{100}{9}=-100+\frac{100}{9}$$

Add -100 to \frac{100}{9}.

$$t^{2}+\frac{20}{3}t+\frac{100}{9}=-\frac{800}{9}$$

Factor t^{2}+\frac{20}{3}t+\frac{100}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{10}{3}\right)^{2}=-\frac{800}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{10}{3}\right)^{2}}=\sqrt{-\frac{800}{9}}$$

Simplify.

$$t+\frac{10}{3}=\frac{20\sqrt{2}i}{3} t+\frac{10}{3}=-\frac{20\sqrt{2}i}{3}$$

Subtract \frac{10}{3} from both sides of the equation.

$$t=\frac{-10+20\sqrt{2}i}{3} t=\frac{-20\sqrt{2}i-10}{3}$$