Solve x^2+7x^2-12x-3=12

$$x ^ { 2 } + 7 x ^ { 2 } - 12 x - 3 = 12$$

Solve for x

$$x = \frac{\sqrt{39} + 3}{4} \approx 2.3112495$$

$$x=\frac{3-\sqrt{39}}{4}\approx -0.8112495$$

Steps Using the Quadratic Formula

Combine x^{2} and 7x^{2} to get 8x^{2}.

$$8x^{2}-12x-3=12$$

Subtract 12 from both sides.

$$8x^{2}-12x-3-12=0$$

Subtract 12 from -3 to get -15.

$$8x^{2}-12x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -12 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 8\left(-15\right)}}{2\times 8}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 8\left(-15\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-12\right)±\sqrt{144-32\left(-15\right)}}{2\times 8}$$

Multiply -32 times -15.

$$x=\frac{-\left(-12\right)±\sqrt{144+480}}{2\times 8}$$

Add 144 to 480.

$$x=\frac{-\left(-12\right)±\sqrt{624}}{2\times 8}$$

Take the square root of 624.

$$x=\frac{-\left(-12\right)±4\sqrt{39}}{2\times 8}$$

The opposite of -12 is 12.

$$x=\frac{12±4\sqrt{39}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{12±4\sqrt{39}}{16}$$

Now solve the equation x=\frac{12±4\sqrt{39}}{16} when ± is plus. Add 12 to 4\sqrt{39}.

$$x=\frac{4\sqrt{39}+12}{16}$$

Divide 12+4\sqrt{39} by 16.

$$x=\frac{\sqrt{39}+3}{4}$$

Now solve the equation x=\frac{12±4\sqrt{39}}{16} when ± is minus. Subtract 4\sqrt{39} from 12.

$$x=\frac{12-4\sqrt{39}}{16}$$

Divide 12-4\sqrt{39} by 16.

$$x=\frac{3-\sqrt{39}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{39}+3}{4} x=\frac{3-\sqrt{39}}{4}$$

Steps for Completing the Square

Combine x^{2} and 7x^{2} to get 8x^{2}.

$$8x^{2}-12x-3=12$$

Add 3 to both sides.

$$8x^{2}-12x=12+3$$

Add 12 and 3 to get 15.

$$8x^{2}-12x=15$$

Divide both sides by 8.

$$\frac{8x^{2}-12x}{8}=\frac{15}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{12}{8}\right)x=\frac{15}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{3}{2}x=\frac{15}{8}$$

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

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{15}{8}+\left(-\frac{3}{4}\right)^{2}$$

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

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{15}{8}+\frac{9}{16}$$

Add \frac{15}{8} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{39}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{39}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{39}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{39}}{4} x-\frac{3}{4}=-\frac{\sqrt{39}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{39}+3}{4} x=\frac{3-\sqrt{39}}{4}$$