Solve x^2+2x-3=-3/4x-3/4

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

Solve for x

$$x=\frac{\sqrt{265}-11}{8}\approx 0.659852575$$

$$x=\frac{-\sqrt{265}-11}{8}\approx -3.409852575$$

Steps Using the Quadratic Formula

Add \frac{3}{4}x to both sides.

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

Combine 2x and \frac{3}{4}x to get \frac{11}{4}x.

$$x^{2}+\frac{11}{4}x-3=-\frac{3}{4}$$

Add \frac{3}{4} to both sides.

$$x^{2}+\frac{11}{4}x-3+\frac{3}{4}=0$$

Add -3 and \frac{3}{4} to get -\frac{9}{4}.

$$x^{2}+\frac{11}{4}x-\frac{9}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, \frac{11}{4} for b, and -\frac{9}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{11}{4}±\sqrt{\left(\frac{11}{4}\right)^{2}-4\left(-\frac{9}{4}\right)}}{2}$$

Square \frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{11}{4}±\sqrt{\frac{121}{16}-4\left(-\frac{9}{4}\right)}}{2}$$

Multiply -4 times -\frac{9}{4}.

$$x=\frac{-\frac{11}{4}±\sqrt{\frac{121}{16}+9}}{2}$$

Add \frac{121}{16} to 9.

$$x=\frac{-\frac{11}{4}±\sqrt{\frac{265}{16}}}{2}$$

Take the square root of \frac{265}{16}.

$$x=\frac{-\frac{11}{4}±\frac{\sqrt{265}}{4}}{2}$$

Now solve the equation x=\frac{-\frac{11}{4}±\frac{\sqrt{265}}{4}}{2} when ± is plus. Add -\frac{11}{4} to \frac{\sqrt{265}}{4}.

$$x=\frac{\sqrt{265}-11}{2\times 4}$$

Divide \frac{-11+\sqrt{265}}{4} by 2.

$$x=\frac{\sqrt{265}-11}{8}$$

Now solve the equation x=\frac{-\frac{11}{4}±\frac{\sqrt{265}}{4}}{2} when ± is minus. Subtract \frac{\sqrt{265}}{4} from -\frac{11}{4}.

$$x=\frac{-\sqrt{265}-11}{2\times 4}$$

Divide \frac{-11-\sqrt{265}}{4} by 2.

$$x=\frac{-\sqrt{265}-11}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{265}-11}{8} x=\frac{-\sqrt{265}-11}{8}$$

Steps for Completing the Square

Add \frac{3}{4}x to both sides.

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

Combine 2x and \frac{3}{4}x to get \frac{11}{4}x.

$$x^{2}+\frac{11}{4}x-3=-\frac{3}{4}$$

Add 3 to both sides.

$$x^{2}+\frac{11}{4}x=-\frac{3}{4}+3$$

Add -\frac{3}{4} and 3 to get \frac{9}{4}.

$$x^{2}+\frac{11}{4}x=\frac{9}{4}$$

Divide \frac{11}{4}, the coefficient of the x term, by 2 to get \frac{11}{8}. Then add the square of \frac{11}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{4}x+\left(\frac{11}{8}\right)^{2}=\frac{9}{4}+\left(\frac{11}{8}\right)^{2}$$

Square \frac{11}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{4}x+\frac{121}{64}=\frac{9}{4}+\frac{121}{64}$$

Add \frac{9}{4} to \frac{121}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{4}x+\frac{121}{64}=\frac{265}{64}$$

Factor x^{2}+\frac{11}{4}x+\frac{121}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{8}\right)^{2}=\frac{265}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{8}\right)^{2}}=\sqrt{\frac{265}{64}}$$

Simplify.

$$x+\frac{11}{8}=\frac{\sqrt{265}}{8} x+\frac{11}{8}=-\frac{\sqrt{265}}{8}$$

Subtract \frac{11}{8} from both sides of the equation.

$$x=\frac{\sqrt{265}-11}{8} x=\frac{-\sqrt{265}-11}{8}$$