Solve 2x^2-78x=48

$$2 x ^ { 2 } - 78 x = 48$$

Solve for x

$$x = \frac{7 \sqrt{33} + 39}{2} \approx 39.605969263$$

$$x=\frac{39-7\sqrt{33}}{2}\approx -0.605969263$$

Steps Using the Quadratic Formula

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.

$$2x^{2}-78x=48$$

Subtract 48 from both sides of the equation.

$$2x^{2}-78x-48=48-48$$

Subtracting 48 from itself leaves 0.

$$2x^{2}-78x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -78 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-78\right)±\sqrt{\left(-78\right)^{2}-4\times 2\left(-48\right)}}{2\times 2}$$

Square -78.

$$x=\frac{-\left(-78\right)±\sqrt{6084-4\times 2\left(-48\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-78\right)±\sqrt{6084-8\left(-48\right)}}{2\times 2}$$

Multiply -8 times -48.

$$x=\frac{-\left(-78\right)±\sqrt{6084+384}}{2\times 2}$$

Add 6084 to 384.

$$x=\frac{-\left(-78\right)±\sqrt{6468}}{2\times 2}$$

Take the square root of 6468.

$$x=\frac{-\left(-78\right)±14\sqrt{33}}{2\times 2}$$

The opposite of -78 is 78.

$$x=\frac{78±14\sqrt{33}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{78±14\sqrt{33}}{4}$$

Now solve the equation x=\frac{78±14\sqrt{33}}{4} when ± is plus. Add 78 to 14\sqrt{33}.

$$x=\frac{14\sqrt{33}+78}{4}$$

Divide 78+14\sqrt{33} by 4.

$$x=\frac{7\sqrt{33}+39}{2}$$

Now solve the equation x=\frac{78±14\sqrt{33}}{4} when ± is minus. Subtract 14\sqrt{33} from 78.

$$x=\frac{78-14\sqrt{33}}{4}$$

Divide 78-14\sqrt{33} by 4.

$$x=\frac{39-7\sqrt{33}}{2}$$

The equation is now solved.

$$x=\frac{7\sqrt{33}+39}{2} x=\frac{39-7\sqrt{33}}{2}$$

Steps for Completing the Square

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.

$$2x^{2}-78x=48$$

Divide both sides by 2.

$$\frac{2x^{2}-78x}{2}=\frac{48}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{78}{2}\right)x=\frac{48}{2}$$

Divide -78 by 2.

$$x^{2}-39x=\frac{48}{2}$$

Divide 48 by 2.

$$x^{2}-39x=24$$

Divide -39, the coefficient of the x term, by 2 to get -\frac{39}{2}. Then add the square of -\frac{39}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-39x+\left(-\frac{39}{2}\right)^{2}=24+\left(-\frac{39}{2}\right)^{2}$$

Square -\frac{39}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-39x+\frac{1521}{4}=24+\frac{1521}{4}$$

Add 24 to \frac{1521}{4}.

$$x^{2}-39x+\frac{1521}{4}=\frac{1617}{4}$$

Factor x^{2}-39x+\frac{1521}{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}.

$$\left(x-\frac{39}{2}\right)^{2}=\frac{1617}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{39}{2}\right)^{2}}=\sqrt{\frac{1617}{4}}$$

Simplify.

$$x-\frac{39}{2}=\frac{7\sqrt{33}}{2} x-\frac{39}{2}=-\frac{7\sqrt{33}}{2}$$

Add \frac{39}{2} to both sides of the equation.

$$x=\frac{7\sqrt{33}+39}{2} x=\frac{39-7\sqrt{33}}{2}$$