Solve 16.8=20h^2+10h

$$16.8=20 { h }^{ 2 } +10h$$

Solve for h

$$h=-1.2$$

$$h=0.7$$

Steps Using the Quadratic Formula

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

$$20h^{2}+10h=16.8$$

Subtract 16.8 from both sides.

$$20h^{2}+10h-16.8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 10 for b, and -16.8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$h=\frac{-10±\sqrt{10^{2}-4\times 20\left(-16.8\right)}}{2\times 20}$$

Square 10.

$$h=\frac{-10±\sqrt{100-4\times 20\left(-16.8\right)}}{2\times 20}$$

Multiply -4 times 20.

$$h=\frac{-10±\sqrt{100-80\left(-16.8\right)}}{2\times 20}$$

Multiply -80 times -16.8.

$$h=\frac{-10±\sqrt{100+1344}}{2\times 20}$$

Add 100 to 1344.

$$h=\frac{-10±\sqrt{1444}}{2\times 20}$$

Take the square root of 1444.

$$h=\frac{-10±38}{2\times 20}$$

Multiply 2 times 20.

$$h=\frac{-10±38}{40}$$

Now solve the equation h=\frac{-10±38}{40} when ± is plus. Add -10 to 38.

$$h=\frac{28}{40}$$

Reduce the fraction \frac{28}{40} to lowest terms by extracting and canceling out 4.

$$h=\frac{7}{10}$$

Now solve the equation h=\frac{-10±38}{40} when ± is minus. Subtract 38 from -10.

$$h=-\frac{48}{40}$$

Reduce the fraction \frac{-48}{40} to lowest terms by extracting and canceling out 8.

$$h=-\frac{6}{5}$$

The equation is now solved.

$$h=\frac{7}{10} h=-\frac{6}{5}$$

Steps for Completing the Square

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

$$20h^{2}+10h=16.8$$

Divide both sides by 20.

$$\frac{20h^{2}+10h}{20}=\frac{16.8}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$h^{2}+\frac{10}{20}h=\frac{16.8}{20}$$

Reduce the fraction \frac{10}{20} to lowest terms by extracting and canceling out 10.

$$h^{2}+\frac{1}{2}h=\frac{16.8}{20}$$

Divide 16.8 by 20.

$$h^{2}+\frac{1}{2}h=0.84$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$h^{2}+\frac{1}{2}h+\left(\frac{1}{4}\right)^{2}=0.84+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$h^{2}+\frac{1}{2}h+\frac{1}{16}=0.84+\frac{1}{16}$$

Add 0.84 to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$h^{2}+\frac{1}{2}h+\frac{1}{16}=\frac{361}{400}$$

Factor h^{2}+\frac{1}{2}h+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(h+\frac{1}{4}\right)^{2}=\frac{361}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(h+\frac{1}{4}\right)^{2}}=\sqrt{\frac{361}{400}}$$

Simplify.

$$h+\frac{1}{4}=\frac{19}{20} h+\frac{1}{4}=-\frac{19}{20}$$

Subtract \frac{1}{4} from both sides of the equation.

$$h=\frac{7}{10} h=-\frac{6}{5}$$