Solve (x)=-3x^2+19x-20

$$(x)=-3 { x }^{ 2 } +19x-20$$

Solve for x

$$x=\frac{\sqrt{21}}{3}+3\approx 4.527525232$$

$$x=-\frac{\sqrt{21}}{3}+3\approx 1.472474768$$

Steps Using the Quadratic Formula

Add 3x^{2} to both sides.

$$x+3x^{2}=19x-20$$

Subtract 19x from both sides.

$$x+3x^{2}-19x=-20$$

Combine x and -19x to get -18x.

$$-18x+3x^{2}=-20$$

Add 20 to both sides.

$$-18x+3x^{2}+20=0$$

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.

$$3x^{2}-18x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -18 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 20}}{2\times 3}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 3\times 20}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-18\right)±\sqrt{324-12\times 20}}{2\times 3}$$

Multiply -12 times 20.

$$x=\frac{-\left(-18\right)±\sqrt{324-240}}{2\times 3}$$

Add 324 to -240.

$$x=\frac{-\left(-18\right)±\sqrt{84}}{2\times 3}$$

Take the square root of 84.

$$x=\frac{-\left(-18\right)±2\sqrt{21}}{2\times 3}$$

The opposite of -18 is 18.

$$x=\frac{18±2\sqrt{21}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{18±2\sqrt{21}}{6}$$

Now solve the equation x=\frac{18±2\sqrt{21}}{6} when ± is plus. Add 18 to 2\sqrt{21}.

$$x=\frac{2\sqrt{21}+18}{6}$$

Divide 18+2\sqrt{21} by 6.

$$x=\frac{\sqrt{21}}{3}+3$$

Now solve the equation x=\frac{18±2\sqrt{21}}{6} when ± is minus. Subtract 2\sqrt{21} from 18.

$$x=\frac{18-2\sqrt{21}}{6}$$

Divide 18-2\sqrt{21} by 6.

$$x=-\frac{\sqrt{21}}{3}+3$$

The equation is now solved.

$$x=\frac{\sqrt{21}}{3}+3 x=-\frac{\sqrt{21}}{3}+3$$

Steps for Completing the Square

Add 3x^{2} to both sides.

$$x+3x^{2}=19x-20$$

Subtract 19x from both sides.

$$x+3x^{2}-19x=-20$$

Combine x and -19x to get -18x.

$$-18x+3x^{2}=-20$$

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.

$$3x^{2}-18x=-20$$

Divide both sides by 3.

$$\frac{3x^{2}-18x}{3}=-\frac{20}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{18}{3}\right)x=-\frac{20}{3}$$

Divide -18 by 3.

$$x^{2}-6x=-\frac{20}{3}$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-\frac{20}{3}+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-\frac{20}{3}+9$$

Add -\frac{20}{3} to 9.

$$x^{2}-6x+9=\frac{7}{3}$$

Factor x^{2}-6x+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}.

$$\left(x-3\right)^{2}=\frac{7}{3}$$

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=\frac{\sqrt{21}}{3} x-3=-\frac{\sqrt{21}}{3}$$

Add 3 to both sides of the equation.

$$x=\frac{\sqrt{21}}{3}+3 x=-\frac{\sqrt{21}}{3}+3$$