Solve 7x^2-3x+x^2-3=5x

$$7 x ^ { 2 } - 3 x + x ^ { 2 } - 3 = 5 x$$

Solve for x

$$x=\frac{\sqrt{10}}{4}+\frac{1}{2}\approx 1.290569415$$

$$x=-\frac{\sqrt{10}}{4}+\frac{1}{2}\approx -0.290569415$$

Steps Using the Quadratic Formula

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

$$8x^{2}-3x-3=5x$$

Subtract 5x from both sides.

$$8x^{2}-3x-3-5x=0$$

Combine -3x and -5x to get -8x.

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

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

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

Square -8.

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

Multiply -4 times 8.

$$x=\frac{-\left(-8\right)±\sqrt{64-32\left(-3\right)}}{2\times 8}$$

Multiply -32 times -3.

$$x=\frac{-\left(-8\right)±\sqrt{64+96}}{2\times 8}$$

Add 64 to 96.

$$x=\frac{-\left(-8\right)±\sqrt{160}}{2\times 8}$$

Take the square root of 160.

$$x=\frac{-\left(-8\right)±4\sqrt{10}}{2\times 8}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{10}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{8±4\sqrt{10}}{16}$$

Now solve the equation x=\frac{8±4\sqrt{10}}{16} when ± is plus. Add 8 to 4\sqrt{10}.

$$x=\frac{4\sqrt{10}+8}{16}$$

Divide 8+4\sqrt{10} by 16.

$$x=\frac{\sqrt{10}}{4}+\frac{1}{2}$$

Now solve the equation x=\frac{8±4\sqrt{10}}{16} when ± is minus. Subtract 4\sqrt{10} from 8.

$$x=\frac{8-4\sqrt{10}}{16}$$

Divide 8-4\sqrt{10} by 16.

$$x=-\frac{\sqrt{10}}{4}+\frac{1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{10}}{4}+\frac{1}{2} x=-\frac{\sqrt{10}}{4}+\frac{1}{2}$$

Steps for Completing the Square

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

$$8x^{2}-3x-3=5x$$

Subtract 5x from both sides.

$$8x^{2}-3x-3-5x=0$$

Combine -3x and -5x to get -8x.

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

Add 3 to both sides. Anything plus zero gives itself.

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

Divide both sides by 8.

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

Dividing by 8 undoes the multiplication by 8.

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

Divide -8 by 8.

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

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{3}{8}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{3}{8}+\frac{1}{4}$$

Add \frac{3}{8} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{5}{8}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{5}{8}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{8}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{10}}{4} x-\frac{1}{2}=-\frac{\sqrt{10}}{4}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{10}}{4}+\frac{1}{2} x=-\frac{\sqrt{10}}{4}+\frac{1}{2}$$