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

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

Solve for x (complex solution)

$$x=\frac{-5+\sqrt{11}i}{2}\approx -2.5+1.658312395i$$

$$x=\frac{-\sqrt{11}i-5}{2}\approx -2.5-1.658312395i$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

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

Combine 5x^{2} and -4x^{2} to get x^{2}.

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

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

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

Add 5 to both sides.

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

Add 4 and 5 to get 9.

$$x^{2}+5x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 9}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 9}}{2}$$

Multiply -4 times 9.

$$x=\frac{-5±\sqrt{25-36}}{2}$$

Add 25 to -36.

$$x=\frac{-5±\sqrt{-11}}{2}$$

Take the square root of -11.

$$x=\frac{-5±\sqrt{11}i}{2}$$

Now solve the equation x=\frac{-5±\sqrt{11}i}{2} when ± is plus. Add -5 to i\sqrt{11}.

$$x=\frac{-5+\sqrt{11}i}{2}$$

Now solve the equation x=\frac{-5±\sqrt{11}i}{2} when ± is minus. Subtract i\sqrt{11} from -5.

$$x=\frac{-\sqrt{11}i-5}{2}$$

The equation is now solved.

$$x=\frac{-5+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i-5}{2}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

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

Combine 5x^{2} and -4x^{2} to get x^{2}.

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

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

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

Subtract 4 from both sides.

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

Subtract 4 from -5 to get -9.

$$x^{2}+5x=-9$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-9+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=-9+\frac{25}{4}$$

Add -9 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=-\frac{11}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=-\frac{11}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{-\frac{11}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{11}i}{2} x+\frac{5}{2}=-\frac{\sqrt{11}i}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{-5+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i-5}{2}$$