Solve 4x^2-7x-3=-9x

$$4 x ^ { 2 } - 7 x - 3 = - 9 x$$

Solve for x

$$x=\frac{\sqrt{13}-1}{4}\approx 0.651387819$$

$$x=\frac{-\sqrt{13}-1}{4}\approx -1.151387819$$

Steps Using the Quadratic Formula

Add 9x to both sides.

$$4x^{2}-7x-3+9x=0$$

Combine -7x and 9x to get 2x.

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

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

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

Square 2.

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

Multiply -4 times 4.

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

Multiply -16 times -3.

$$x=\frac{-2±\sqrt{4+48}}{2\times 4}$$

Add 4 to 48.

$$x=\frac{-2±\sqrt{52}}{2\times 4}$$

Take the square root of 52.

$$x=\frac{-2±2\sqrt{13}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-2±2\sqrt{13}}{8}$$

Now solve the equation x=\frac{-2±2\sqrt{13}}{8} when ± is plus. Add -2 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}-2}{8}$$

Divide -2+2\sqrt{13} by 8.

$$x=\frac{\sqrt{13}-1}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{13}}{8} when ± is minus. Subtract 2\sqrt{13} from -2.

$$x=\frac{-2\sqrt{13}-2}{8}$$

Divide -2-2\sqrt{13} by 8.

$$x=\frac{-\sqrt{13}-1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{13}-1}{4} x=\frac{-\sqrt{13}-1}{4}$$

Steps for Completing the Square

Add 9x to both sides.

$$4x^{2}-7x-3+9x=0$$

Combine -7x and 9x to get 2x.

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

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

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{3}{4}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{3}{4}+\frac{1}{16}$$

Add \frac{3}{4} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{13}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{13}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{13}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{13}}{4} x+\frac{1}{4}=-\frac{\sqrt{13}}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{13}-1}{4} x=\frac{-\sqrt{13}-1}{4}$$