Solve 4x(5x+6)=7-3x

$$4x(5x+6)=7-3x$$

Solve for x

$$x=\frac{\sqrt{1289}-27}{40}\approx 0.222566154$$

$$x=\frac{-\sqrt{1289}-27}{40}\approx -1.572566154$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4x by 5x+6.

$$20x^{2}+24x=7-3x$$

Subtract 7 from both sides.

$$20x^{2}+24x-7=-3x$$

Add 3x to both sides.

$$20x^{2}+24x-7+3x=0$$

Combine 24x and 3x to get 27x.

$$20x^{2}+27x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 27 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-27±\sqrt{27^{2}-4\times 20\left(-7\right)}}{2\times 20}$$

Square 27.

$$x=\frac{-27±\sqrt{729-4\times 20\left(-7\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-27±\sqrt{729-80\left(-7\right)}}{2\times 20}$$

Multiply -80 times -7.

$$x=\frac{-27±\sqrt{729+560}}{2\times 20}$$

Add 729 to 560.

$$x=\frac{-27±\sqrt{1289}}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-27±\sqrt{1289}}{40}$$

Now solve the equation x=\frac{-27±\sqrt{1289}}{40} when ± is plus. Add -27 to \sqrt{1289}.

$$x=\frac{\sqrt{1289}-27}{40}$$

Now solve the equation x=\frac{-27±\sqrt{1289}}{40} when ± is minus. Subtract \sqrt{1289} from -27.

$$x=\frac{-\sqrt{1289}-27}{40}$$

The equation is now solved.

$$x=\frac{\sqrt{1289}-27}{40} x=\frac{-\sqrt{1289}-27}{40}$$

Steps for Completing the Square

Use the distributive property to multiply 4x by 5x+6.

$$20x^{2}+24x=7-3x$$

Add 3x to both sides.

$$20x^{2}+24x+3x=7$$

Combine 24x and 3x to get 27x.

$$20x^{2}+27x=7$$

Divide both sides by 20.

$$\frac{20x^{2}+27x}{20}=\frac{7}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\frac{27}{20}x=\frac{7}{20}$$

Divide \frac{27}{20}, the coefficient of the x term, by 2 to get \frac{27}{40}. Then add the square of \frac{27}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{27}{20}x+\left(\frac{27}{40}\right)^{2}=\frac{7}{20}+\left(\frac{27}{40}\right)^{2}$$

Square \frac{27}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{27}{20}x+\frac{729}{1600}=\frac{7}{20}+\frac{729}{1600}$$

Add \frac{7}{20} to \frac{729}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{27}{20}x+\frac{729}{1600}=\frac{1289}{1600}$$

Factor x^{2}+\frac{27}{20}x+\frac{729}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{27}{40}\right)^{2}=\frac{1289}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{27}{40}\right)^{2}}=\sqrt{\frac{1289}{1600}}$$

Simplify.

$$x+\frac{27}{40}=\frac{\sqrt{1289}}{40} x+\frac{27}{40}=-\frac{\sqrt{1289}}{40}$$

Subtract \frac{27}{40} from both sides of the equation.

$$x=\frac{\sqrt{1289}-27}{40} x=\frac{-\sqrt{1289}-27}{40}$$