Solve -39+(2x-3)^2=2(-10)

$$- 39 + ( 2 x - 3 ) ^ { 2 } = 2 ( - 10 )$$

Solve for x

$$x = \frac{\sqrt{19} + 3}{2} \approx 3.679449472$$

$$x=\frac{3-\sqrt{19}}{2}\approx -0.679449472$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$-39+4x^{2}-12x+9=2\left(-10\right)$$

Add -39 and 9 to get -30.

$$-30+4x^{2}-12x=2\left(-10\right)$$

Multiply 2 and -10 to get -20.

$$-30+4x^{2}-12x=-20$$

Add 20 to both sides.

$$-30+4x^{2}-12x+20=0$$

Add -30 and 20 to get -10.

$$-10+4x^{2}-12x=0$$

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.

$$4x^{2}-12x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-10\right)}}{2\times 4}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\left(-10\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\left(-10\right)}}{2\times 4}$$

Multiply -16 times -10.

$$x=\frac{-\left(-12\right)±\sqrt{144+160}}{2\times 4}$$

Add 144 to 160.

$$x=\frac{-\left(-12\right)±\sqrt{304}}{2\times 4}$$

Take the square root of 304.

$$x=\frac{-\left(-12\right)±4\sqrt{19}}{2\times 4}$$

The opposite of -12 is 12.

$$x=\frac{12±4\sqrt{19}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{12±4\sqrt{19}}{8}$$

Now solve the equation x=\frac{12±4\sqrt{19}}{8} when ± is plus. Add 12 to 4\sqrt{19}.

$$x=\frac{4\sqrt{19}+12}{8}$$

Divide 12+4\sqrt{19} by 8.

$$x=\frac{\sqrt{19}+3}{2}$$

Now solve the equation x=\frac{12±4\sqrt{19}}{8} when ± is minus. Subtract 4\sqrt{19} from 12.

$$x=\frac{12-4\sqrt{19}}{8}$$

Divide 12-4\sqrt{19} by 8.

$$x=\frac{3-\sqrt{19}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{19}+3}{2} x=\frac{3-\sqrt{19}}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$-39+4x^{2}-12x+9=2\left(-10\right)$$

Add -39 and 9 to get -30.

$$-30+4x^{2}-12x=2\left(-10\right)$$

Multiply 2 and -10 to get -20.

$$-30+4x^{2}-12x=-20$$

Add 30 to both sides.

$$4x^{2}-12x=-20+30$$

Add -20 and 30 to get 10.

$$4x^{2}-12x=10$$

Divide both sides by 4.

$$\frac{4x^{2}-12x}{4}=\frac{10}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{12}{4}\right)x=\frac{10}{4}$$

Divide -12 by 4.

$$x^{2}-3x=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-3x=\frac{5}{2}$$

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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\frac{5}{2}+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=\frac{5}{2}+\frac{9}{4}$$

Add \frac{5}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=\frac{19}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{19}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{19}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{19}}{2} x-\frac{3}{2}=-\frac{\sqrt{19}}{2}$$

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

$$x=\frac{\sqrt{19}+3}{2} x=\frac{3-\sqrt{19}}{2}$$