Solve 79=4x^2+x+7

$$79 = 4 x ^ { 2 } + x + 7$$

Solve for x

$$x = \frac{\sqrt{1153} - 1}{8} \approx 4.119481712$$

$$x=\frac{-\sqrt{1153}-1}{8}\approx -4.369481712$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+x+7=79$$

Subtract 79 from both sides.

$$4x^{2}+x+7-79=0$$

Subtract 79 from 7 to get -72.

$$4x^{2}+x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 1 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\left(-72\right)}}{2\times 4}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 4\left(-72\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-1±\sqrt{1-16\left(-72\right)}}{2\times 4}$$

Multiply -16 times -72.

$$x=\frac{-1±\sqrt{1+1152}}{2\times 4}$$

Add 1 to 1152.

$$x=\frac{-1±\sqrt{1153}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-1±\sqrt{1153}}{8}$$

Now solve the equation x=\frac{-1±\sqrt{1153}}{8} when ± is plus. Add -1 to \sqrt{1153}.

$$x=\frac{\sqrt{1153}-1}{8}$$

Now solve the equation x=\frac{-1±\sqrt{1153}}{8} when ± is minus. Subtract \sqrt{1153} from -1.

$$x=\frac{-\sqrt{1153}-1}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{1153}-1}{8} x=\frac{-\sqrt{1153}-1}{8}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+x+7=79$$

Subtract 7 from both sides.

$$4x^{2}+x=79-7$$

Subtract 7 from 79 to get 72.

$$4x^{2}+x=72$$

Divide both sides by 4.

$$\frac{4x^{2}+x}{4}=\frac{72}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{1}{4}x=\frac{72}{4}$$

Divide 72 by 4.

$$x^{2}+\frac{1}{4}x=18$$

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.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=18+\left(\frac{1}{8}\right)^{2}$$

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

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=18+\frac{1}{64}$$

Add 18 to \frac{1}{64}.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{1153}{64}$$

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}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{1153}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{1153}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{\sqrt{1153}}{8} x+\frac{1}{8}=-\frac{\sqrt{1153}}{8}$$

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

$$x=\frac{\sqrt{1153}-1}{8} x=\frac{-\sqrt{1153}-1}{8}$$