Solve 64+16q+25q^2=(8+10q)^2

$$64 + 16 q + 25 q ^ { 2 } = ( 8 + 10 q ) ^ { 2 }$$

Solve for q

$$q = -\frac{48}{25} = -1\frac{23}{25} = -1.92$$

$$q=0$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8+10q\right)^{2}.

$$64+16q+25q^{2}=64+160q+100q^{2}$$

Subtract 64 from both sides.

$$64+16q+25q^{2}-64=160q+100q^{2}$$

Subtract 64 from 64 to get 0.

$$16q+25q^{2}=160q+100q^{2}$$

Subtract 160q from both sides.

$$16q+25q^{2}-160q=100q^{2}$$

Combine 16q and -160q to get -144q.

$$-144q+25q^{2}=100q^{2}$$

Subtract 100q^{2} from both sides.

$$-144q+25q^{2}-100q^{2}=0$$

Combine 25q^{2} and -100q^{2} to get -75q^{2}.

$$-144q-75q^{2}=0$$

Factor out q.

$$q\left(-144-75q\right)=0$$

To find equation solutions, solve q=0 and -144-75q=0.

$$q=0 q=-\frac{48}{25}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8+10q\right)^{2}.

$$64+16q+25q^{2}=64+160q+100q^{2}$$

Subtract 64 from both sides.

$$64+16q+25q^{2}-64=160q+100q^{2}$$

Subtract 64 from 64 to get 0.

$$16q+25q^{2}=160q+100q^{2}$$

Subtract 160q from both sides.

$$16q+25q^{2}-160q=100q^{2}$$

Combine 16q and -160q to get -144q.

$$-144q+25q^{2}=100q^{2}$$

Subtract 100q^{2} from both sides.

$$-144q+25q^{2}-100q^{2}=0$$

Combine 25q^{2} and -100q^{2} to get -75q^{2}.

$$-144q-75q^{2}=0$$

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.

$$-75q^{2}-144q=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -75 for a, -144 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-144\right)±\sqrt{\left(-144\right)^{2}}}{2\left(-75\right)}$$

Take the square root of \left(-144\right)^{2}.

$$q=\frac{-\left(-144\right)±144}{2\left(-75\right)}$$

The opposite of -144 is 144.

$$q=\frac{144±144}{2\left(-75\right)}$$

Multiply 2 times -75.

$$q=\frac{144±144}{-150}$$

Now solve the equation q=\frac{144±144}{-150} when ± is plus. Add 144 to 144.

$$q=\frac{288}{-150}$$

Reduce the fraction \frac{288}{-150} to lowest terms by extracting and canceling out 6.

$$q=-\frac{48}{25}$$

Now solve the equation q=\frac{144±144}{-150} when ± is minus. Subtract 144 from 144.

$$q=\frac{0}{-150}$$

Divide 0 by -150.

$$q=0$$

The equation is now solved.

$$q=-\frac{48}{25} q=0$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8+10q\right)^{2}.

$$64+16q+25q^{2}=64+160q+100q^{2}$$

Subtract 160q from both sides.

$$64+16q+25q^{2}-160q=64+100q^{2}$$

Combine 16q and -160q to get -144q.

$$64-144q+25q^{2}=64+100q^{2}$$

Subtract 100q^{2} from both sides.

$$64-144q+25q^{2}-100q^{2}=64$$

Combine 25q^{2} and -100q^{2} to get -75q^{2}.

$$64-144q-75q^{2}=64$$

Subtract 64 from both sides.

$$-144q-75q^{2}=64-64$$

Subtract 64 from 64 to get 0.

$$-144q-75q^{2}=0$$

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.

$$-75q^{2}-144q=0$$

Divide both sides by -75.

$$\frac{-75q^{2}-144q}{-75}=\frac{0}{-75}$$

Dividing by -75 undoes the multiplication by -75.

$$q^{2}+\left(-\frac{144}{-75}\right)q=\frac{0}{-75}$$

Reduce the fraction \frac{-144}{-75} to lowest terms by extracting and canceling out 3.

$$q^{2}+\frac{48}{25}q=\frac{0}{-75}$$

Divide 0 by -75.

$$q^{2}+\frac{48}{25}q=0$$

Divide \frac{48}{25}, the coefficient of the x term, by 2 to get \frac{24}{25}. Then add the square of \frac{24}{25} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}+\frac{48}{25}q+\left(\frac{24}{25}\right)^{2}=\left(\frac{24}{25}\right)^{2}$$

Square \frac{24}{25} by squaring both the numerator and the denominator of the fraction.

$$q^{2}+\frac{48}{25}q+\frac{576}{625}=\frac{576}{625}$$

Factor q^{2}+\frac{48}{25}q+\frac{576}{625}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q+\frac{24}{25}\right)^{2}=\frac{576}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q+\frac{24}{25}\right)^{2}}=\sqrt{\frac{576}{625}}$$

Simplify.

$$q+\frac{24}{25}=\frac{24}{25} q+\frac{24}{25}=-\frac{24}{25}$$

Subtract \frac{24}{25} from both sides of the equation.

$$q=0 q=-\frac{48}{25}$$