Solve 3x+4=296+x^2

$$3x+4 = 296+ { x }^{ 2 }$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{1159}i+3}{2}\approx 1.5-17.022044531i$$

$$x=\frac{3+\sqrt{1159}i}{2}\approx 1.5+17.022044531i$$

Steps Using the Quadratic Formula

Subtract 296 from both sides.

$$3x+4-296=x^{2}$$

Subtract 296 from 4 to get -292.

$$3x-292=x^{2}$$

Subtract x^{2} from both sides.

$$3x-292-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+3x-292=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and -292 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\left(-292\right)}}{2\left(-1\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-292\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\left(-292\right)}}{2\left(-1\right)}$$

Multiply 4 times -292.

$$x=\frac{-3±\sqrt{9-1168}}{2\left(-1\right)}$$

Add 9 to -1168.

$$x=\frac{-3±\sqrt{-1159}}{2\left(-1\right)}$$

Take the square root of -1159.

$$x=\frac{-3±\sqrt{1159}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±\sqrt{1159}i}{-2}$$

Now solve the equation x=\frac{-3±\sqrt{1159}i}{-2} when ± is plus. Add -3 to i\sqrt{1159}.

$$x=\frac{-3+\sqrt{1159}i}{-2}$$

Divide -3+i\sqrt{1159} by -2.

$$x=\frac{-\sqrt{1159}i+3}{2}$$

Now solve the equation x=\frac{-3±\sqrt{1159}i}{-2} when ± is minus. Subtract i\sqrt{1159} from -3.

$$x=\frac{-\sqrt{1159}i-3}{-2}$$

Divide -3-i\sqrt{1159} by -2.

$$x=\frac{3+\sqrt{1159}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{1159}i+3}{2} x=\frac{3+\sqrt{1159}i}{2}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$3x+4-x^{2}=296$$

Subtract 4 from both sides.

$$3x-x^{2}=296-4$$

Subtract 4 from 296 to get 292.

$$3x-x^{2}=292$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+3x=292$$

Divide both sides by -1.

$$\frac{-x^{2}+3x}{-1}=\frac{292}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{3}{-1}x=\frac{292}{-1}$$

Divide 3 by -1.

$$x^{2}-3x=\frac{292}{-1}$$

Divide 292 by -1.

$$x^{2}-3x=-292$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-292+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-292+\frac{9}{4}$$

Add -292 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=-\frac{1159}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{1159}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{1159}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{1159}i}{2} x-\frac{3}{2}=-\frac{\sqrt{1159}i}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+\sqrt{1159}i}{2} x=\frac{-\sqrt{1159}i+3}{2}$$