Solve x^2-4(x^2-5x+6)=0

$$x ^ { 2 } - 4 ( x ^ { 2 } - 5 x + 6 ) = 0$$

Solve for x

$$x = \frac{2 \sqrt{7} + 10}{3} \approx 5.097167541$$

$$x = \frac{10 - 2 \sqrt{7}}{3} \approx 1.569499126$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -4 by x^{2}-5x+6.

$$x^{2}-4x^{2}+20x-24=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+20x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 20 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-3\right)\left(-24\right)}}{2\left(-3\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-3\right)\left(-24\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-20±\sqrt{400+12\left(-24\right)}}{2\left(-3\right)}$$

Multiply 12 times -24.

$$x=\frac{-20±\sqrt{400-288}}{2\left(-3\right)}$$

Add 400 to -288.

$$x=\frac{-20±\sqrt{112}}{2\left(-3\right)}$$

Take the square root of 112.

$$x=\frac{-20±4\sqrt{7}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-20±4\sqrt{7}}{-6}$$

Now solve the equation x=\frac{-20±4\sqrt{7}}{-6} when ± is plus. Add -20 to 4\sqrt{7}.

$$x=\frac{4\sqrt{7}-20}{-6}$$

Divide -20+4\sqrt{7} by -6.

$$x=\frac{10-2\sqrt{7}}{3}$$

Now solve the equation x=\frac{-20±4\sqrt{7}}{-6} when ± is minus. Subtract 4\sqrt{7} from -20.

$$x=\frac{-4\sqrt{7}-20}{-6}$$

Divide -20-4\sqrt{7} by -6.

$$x=\frac{2\sqrt{7}+10}{3}$$

The equation is now solved.

$$x=\frac{10-2\sqrt{7}}{3} x=\frac{2\sqrt{7}+10}{3}$$

Steps for Completing the Square

Use the distributive property to multiply -4 by x^{2}-5x+6.

$$x^{2}-4x^{2}+20x-24=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+20x-24=0$$

Add 24 to both sides. Anything plus zero gives itself.

$$-3x^{2}+20x=24$$

Divide both sides by -3.

$$\frac{-3x^{2}+20x}{-3}=\frac{24}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{20}{-3}x=\frac{24}{-3}$$

Divide 20 by -3.

$$x^{2}-\frac{20}{3}x=\frac{24}{-3}$$

Divide 24 by -3.

$$x^{2}-\frac{20}{3}x=-8$$

Divide -\frac{20}{3}, the coefficient of the x term, by 2 to get -\frac{10}{3}. Then add the square of -\frac{10}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{20}{3}x+\left(-\frac{10}{3}\right)^{2}=-8+\left(-\frac{10}{3}\right)^{2}$$

Square -\frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=-8+\frac{100}{9}$$

Add -8 to \frac{100}{9}.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{28}{9}$$

Factor x^{2}-\frac{20}{3}x+\frac{100}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{10}{3}\right)^{2}=\frac{28}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{3}\right)^{2}}=\sqrt{\frac{28}{9}}$$

Simplify.

$$x-\frac{10}{3}=\frac{2\sqrt{7}}{3} x-\frac{10}{3}=-\frac{2\sqrt{7}}{3}$$

Add \frac{10}{3} to both sides of the equation.

$$x=\frac{2\sqrt{7}+10}{3} x=\frac{10-2\sqrt{7}}{3}$$