Solve 4x^2+4x-3=3/2x-5

$$4 { x }^{ 2 } +4x-3 = \frac{ 3 }{ 2 } x-5$$

Solve for x (complex solution)

$$x=\frac{-5+\sqrt{103}i}{16}\approx -0.3125+0.634305723i$$

$$x=\frac{-\sqrt{103}i-5}{16}\approx -0.3125-0.634305723i$$

Steps Using the Quadratic Formula

Subtract \frac{3}{2}x from both sides.

$$4x^{2}+4x-3-\frac{3}{2}x=-5$$

Combine 4x and -\frac{3}{2}x to get \frac{5}{2}x.

$$4x^{2}+\frac{5}{2}x-3=-5$$

Add 5 to both sides.

$$4x^{2}+\frac{5}{2}x-3+5=0$$

Add -3 and 5 to get 2.

$$4x^{2}+\frac{5}{2}x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, \frac{5}{2} for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{5}{2}±\sqrt{\left(\frac{5}{2}\right)^{2}-4\times 4\times 2}}{2\times 4}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-4\times 4\times 2}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-16\times 2}}{2\times 4}$$

Multiply -16 times 2.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-32}}{2\times 4}$$

Add \frac{25}{4} to -32.

$$x=\frac{-\frac{5}{2}±\sqrt{-\frac{103}{4}}}{2\times 4}$$

Take the square root of -\frac{103}{4}.

$$x=\frac{-\frac{5}{2}±\frac{\sqrt{103}i}{2}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-\frac{5}{2}±\frac{\sqrt{103}i}{2}}{8}$$

Now solve the equation x=\frac{-\frac{5}{2}±\frac{\sqrt{103}i}{2}}{8} when ± is plus. Add -\frac{5}{2} to \frac{i\sqrt{103}}{2}.

$$x=\frac{-5+\sqrt{103}i}{2\times 8}$$

Divide \frac{-5+i\sqrt{103}}{2} by 8.

$$x=\frac{-5+\sqrt{103}i}{16}$$

Now solve the equation x=\frac{-\frac{5}{2}±\frac{\sqrt{103}i}{2}}{8} when ± is minus. Subtract \frac{i\sqrt{103}}{2} from -\frac{5}{2}.

$$x=\frac{-\sqrt{103}i-5}{2\times 8}$$

Divide \frac{-5-i\sqrt{103}}{2} by 8.

$$x=\frac{-\sqrt{103}i-5}{16}$$

The equation is now solved.

$$x=\frac{-5+\sqrt{103}i}{16} x=\frac{-\sqrt{103}i-5}{16}$$

Steps for Completing the Square

Subtract \frac{3}{2}x from both sides.

$$4x^{2}+4x-3-\frac{3}{2}x=-5$$

Combine 4x and -\frac{3}{2}x to get \frac{5}{2}x.

$$4x^{2}+\frac{5}{2}x-3=-5$$

Add 3 to both sides.

$$4x^{2}+\frac{5}{2}x=-5+3$$

Add -5 and 3 to get -2.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide \frac{5}{2} by 4.

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

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

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

Divide \frac{5}{8}, the coefficient of the x term, by 2 to get \frac{5}{16}. Then add the square of \frac{5}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{8}x+\left(\frac{5}{16}\right)^{2}=-\frac{1}{2}+\left(\frac{5}{16}\right)^{2}$$

Square \frac{5}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{8}x+\frac{25}{256}=-\frac{1}{2}+\frac{25}{256}$$

Add -\frac{1}{2} to \frac{25}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{8}x+\frac{25}{256}=-\frac{103}{256}$$

Factor x^{2}+\frac{5}{8}x+\frac{25}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{16}\right)^{2}=-\frac{103}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{16}\right)^{2}}=\sqrt{-\frac{103}{256}}$$

Simplify.

$$x+\frac{5}{16}=\frac{\sqrt{103}i}{16} x+\frac{5}{16}=-\frac{\sqrt{103}i}{16}$$

Subtract \frac{5}{16} from both sides of the equation.

$$x=\frac{-5+\sqrt{103}i}{16} x=\frac{-\sqrt{103}i-5}{16}$$