Solve 3x^2=30x+600

$$3 x ^ { 2 } = 30 x + 600$$

Solve for x

$$x=-10$$

$$x=20$$

Steps Using Factoring By Grouping

Subtract 30x from both sides.

$$3x^{2}-30x=600$$

Subtract 600 from both sides.

$$3x^{2}-30x-600=0$$

Divide both sides by 3.

$$x^{2}-10x-200=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-200. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-200\right)=-200$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -200.

$$1,-200 2,-100 4,-50 5,-40 8,-25 10,-20$$

Calculate the sum for each pair.

$$1-200=-199 2-100=-98 4-50=-46 5-40=-35 8-25=-17 10-20=-10$$

The solution is the pair that gives sum -10.

$$a=-20 b=10$$

Rewrite x^{2}-10x-200 as \left(x^{2}-20x\right)+\left(10x-200\right).

$$\left(x^{2}-20x\right)+\left(10x-200\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(x-20\right)+10\left(x-20\right)$$

Factor out common term x-20 by using distributive property.

$$\left(x-20\right)\left(x+10\right)$$

To find equation solutions, solve x-20=0 and x+10=0.

$$x=20 x=-10$$

Steps Using the Quadratic Formula

Subtract 30x from both sides.

$$3x^{2}-30x=600$$

Subtract 600 from both sides.

$$3x^{2}-30x-600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -30 for b, and -600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 3\left(-600\right)}}{2\times 3}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 3\left(-600\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-30\right)±\sqrt{900-12\left(-600\right)}}{2\times 3}$$

Multiply -12 times -600.

$$x=\frac{-\left(-30\right)±\sqrt{900+7200}}{2\times 3}$$

Add 900 to 7200.

$$x=\frac{-\left(-30\right)±\sqrt{8100}}{2\times 3}$$

Take the square root of 8100.

$$x=\frac{-\left(-30\right)±90}{2\times 3}$$

The opposite of -30 is 30.

$$x=\frac{30±90}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{30±90}{6}$$

Now solve the equation x=\frac{30±90}{6} when ± is plus. Add 30 to 90.

$$x=\frac{120}{6}$$

Divide 120 by 6.

$$x=20$$

Now solve the equation x=\frac{30±90}{6} when ± is minus. Subtract 90 from 30.

$$x=-\frac{60}{6}$$

Divide -60 by 6.

$$x=-10$$

The equation is now solved.

$$x=20 x=-10$$

Steps for Completing the Square

Subtract 30x from both sides.

$$3x^{2}-30x=600$$

Divide both sides by 3.

$$\frac{3x^{2}-30x}{3}=\frac{600}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{30}{3}\right)x=\frac{600}{3}$$

Divide -30 by 3.

$$x^{2}-10x=\frac{600}{3}$$

Divide 600 by 3.

$$x^{2}-10x=200$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=200+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=200+25$$

Add 200 to 25.

$$x^{2}-10x+25=225$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{225}$$

Simplify.

$$x-5=15 x-5=-15$$

Add 5 to both sides of the equation.

$$x=20 x=-10$$