Solve x^2-4x^2-8x+7=0

$$x ^ { 2 } - 4 x ^ { 2 } - 8 x + 7 = 0$$

Solve for x

$$x=\frac{\sqrt{37}-4}{3}\approx 0.694254177$$

$$x=\frac{-\sqrt{37}-4}{3}\approx -3.360920843$$

Steps Using the Quadratic Formula

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-8x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -8 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-3\right)\times 7}}{2\left(-3\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-3\right)\times 7}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-8\right)±\sqrt{64+12\times 7}}{2\left(-3\right)}$$

Multiply 12 times 7.

$$x=\frac{-\left(-8\right)±\sqrt{64+84}}{2\left(-3\right)}$$

Add 64 to 84.

$$x=\frac{-\left(-8\right)±\sqrt{148}}{2\left(-3\right)}$$

Take the square root of 148.

$$x=\frac{-\left(-8\right)±2\sqrt{37}}{2\left(-3\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{37}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{8±2\sqrt{37}}{-6}$$

Now solve the equation x=\frac{8±2\sqrt{37}}{-6} when ± is plus. Add 8 to 2\sqrt{37}.

$$x=\frac{2\sqrt{37}+8}{-6}$$

Divide 8+2\sqrt{37} by -6.

$$x=\frac{-\sqrt{37}-4}{3}$$

Now solve the equation x=\frac{8±2\sqrt{37}}{-6} when ± is minus. Subtract 2\sqrt{37} from 8.

$$x=\frac{8-2\sqrt{37}}{-6}$$

Divide 8-2\sqrt{37} by -6.

$$x=\frac{\sqrt{37}-4}{3}$$

The equation is now solved.

$$x=\frac{-\sqrt{37}-4}{3} x=\frac{\sqrt{37}-4}{3}$$

Steps for Completing the Square

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-8x+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$-3x^{2}-8x=-7$$

Divide both sides by -3.

$$\frac{-3x^{2}-8x}{-3}=-\frac{7}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{8}{-3}\right)x=-\frac{7}{-3}$$

Divide -8 by -3.

$$x^{2}+\frac{8}{3}x=-\frac{7}{-3}$$

Divide -7 by -3.

$$x^{2}+\frac{8}{3}x=\frac{7}{3}$$

Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=\frac{7}{3}+\left(\frac{4}{3}\right)^{2}$$

Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{7}{3}+\frac{16}{9}$$

Add \frac{7}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{37}{9}$$

Factor x^{2}+\frac{8}{3}x+\frac{16}{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}.

$$\left(x+\frac{4}{3}\right)^{2}=\frac{37}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{3}\right)^{2}}=\sqrt{\frac{37}{9}}$$

Simplify.

$$x+\frac{4}{3}=\frac{\sqrt{37}}{3} x+\frac{4}{3}=-\frac{\sqrt{37}}{3}$$

Subtract \frac{4}{3} from both sides of the equation.

$$x=\frac{\sqrt{37}-4}{3} x=\frac{-\sqrt{37}-4}{3}$$