Solve x^2+x^2+6x-3x+10=15-1

$$x ^ { 2 } + x ^ { 2 } + 6 x - 3 x + 10 = 15 - 1$$

Solve for x

$$x=\frac{\sqrt{41}-3}{4}\approx 0.850781059$$

$$x=\frac{-\sqrt{41}-3}{4}\approx -2.350781059$$

Steps Using the Quadratic Formula

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

$$2x^{2}+6x-3x+10=15-1$$

Combine 6x and -3x to get 3x.

$$2x^{2}+3x+10=15-1$$

Subtract 1 from 15 to get 14.

$$2x^{2}+3x+10=14$$

Subtract 14 from both sides.

$$2x^{2}+3x+10-14=0$$

Subtract 14 from 10 to get -4.

$$2x^{2}+3x-4=0$$

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

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-4\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$x=\frac{-3±\sqrt{9+32}}{2\times 2}$$

Add 9 to 32.

$$x=\frac{-3±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±\sqrt{41}}{4}$$

Now solve the equation x=\frac{-3±\sqrt{41}}{4} when ± is plus. Add -3 to \sqrt{41}.

$$x=\frac{\sqrt{41}-3}{4}$$

Now solve the equation x=\frac{-3±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from -3.

$$x=\frac{-\sqrt{41}-3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$

Steps for Completing the Square

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

$$2x^{2}+6x-3x+10=15-1$$

Combine 6x and -3x to get 3x.

$$2x^{2}+3x+10=15-1$$

Subtract 1 from 15 to get 14.

$$2x^{2}+3x+10=14$$

Subtract 10 from both sides.

$$2x^{2}+3x=14-10$$

Subtract 10 from 14 to get 4.

$$2x^{2}+3x=4$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 4 by 2.

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

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.

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

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

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

Add 2 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{41}{16}$$

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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{41}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{41}}{4} x+\frac{3}{4}=-\frac{\sqrt{41}}{4}$$

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

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$