Solve 4x^2+x-112=014

$$4 x ^ { 2 } + x - 112 = 014$$

Solve for x

$$x = \frac{\sqrt{1793} - 1}{8} \approx 5.16797884$$

$$x=\frac{-\sqrt{1793}-1}{8}\approx -5.41797884$$

Steps Using the Quadratic Formula

Multiply 0 and 14 to get 0.

$$4x^{2}+x-112=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 1 for b, and -112 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\left(-112\right)}}{2\times 4}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 4\left(-112\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-1±\sqrt{1-16\left(-112\right)}}{2\times 4}$$

Multiply -16 times -112.

$$x=\frac{-1±\sqrt{1+1792}}{2\times 4}$$

Add 1 to 1792.

$$x=\frac{-1±\sqrt{1793}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-1±\sqrt{1793}}{8}$$

Now solve the equation x=\frac{-1±\sqrt{1793}}{8} when ± is plus. Add -1 to \sqrt{1793}.

$$x=\frac{\sqrt{1793}-1}{8}$$

Now solve the equation x=\frac{-1±\sqrt{1793}}{8} when ± is minus. Subtract \sqrt{1793} from -1.

$$x=\frac{-\sqrt{1793}-1}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{1793}-1}{8} x=\frac{-\sqrt{1793}-1}{8}$$

Steps for Completing the Square

Multiply 0 and 14 to get 0.

$$4x^{2}+x-112=0$$

Add 112 to both sides. Anything plus zero gives itself.

$$4x^{2}+x=112$$

Divide both sides by 4.

$$\frac{4x^{2}+x}{4}=\frac{112}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{1}{4}x=\frac{112}{4}$$

Divide 112 by 4.

$$x^{2}+\frac{1}{4}x=28$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=28+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=28+\frac{1}{64}$$

Add 28 to \frac{1}{64}.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{1793}{64}$$

Factor x^{2}+\frac{1}{4}x+\frac{1}{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}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{1793}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{1793}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{\sqrt{1793}}{8} x+\frac{1}{8}=-\frac{\sqrt{1793}}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$x=\frac{\sqrt{1793}-1}{8} x=\frac{-\sqrt{1793}-1}{8}$$