Solve 0=-5left(x^2-10x-7200right)

$$0=-5 \left( { x }^{ 2 } -10x-7200 \right)$$

Solve for x

$$x=-80$$

$$x=90$$

Steps Using Factoring

Divide both sides by -5. Zero divided by any non-zero number gives zero.

$$0=x^{2}-10x-7200$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-10x-7200=0$$

To solve the equation, factor x^{2}-10x-7200 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=-7200$$

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 -7200.

$$1,-7200 2,-3600 3,-2400 4,-1800 5,-1440 6,-1200 8,-900 9,-800 10,-720 12,-600 15,-480 16,-450 18,-400 20,-360 24,-300 25,-288 30,-240 32,-225 36,-200 40,-180 45,-160 48,-150 50,-144 60,-120 72,-100 75,-96 80,-90$$

Calculate the sum for each pair.

$$1-7200=-7199 2-3600=-3598 3-2400=-2397 4-1800=-1796 5-1440=-1435 6-1200=-1194 8-900=-892 9-800=-791 10-720=-710 12-600=-588 15-480=-465 16-450=-434 18-400=-382 20-360=-340 24-300=-276 25-288=-263 30-240=-210 32-225=-193 36-200=-164 40-180=-140 45-160=-115 48-150=-102 50-144=-94 60-120=-60 72-100=-28 75-96=-21 80-90=-10$$

The solution is the pair that gives sum -10.

$$a=-90 b=80$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-90\right)\left(x+80\right)$$

To find equation solutions, solve x-90=0 and x+80=0.

$$x=90 x=-80$$

Steps Using Factoring By Grouping

Divide both sides by -5. Zero divided by any non-zero number gives zero.

$$0=x^{2}-10x-7200$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-10x-7200=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-7200. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-7200\right)=-7200$$

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 -7200.

$$1,-7200 2,-3600 3,-2400 4,-1800 5,-1440 6,-1200 8,-900 9,-800 10,-720 12,-600 15,-480 16,-450 18,-400 20,-360 24,-300 25,-288 30,-240 32,-225 36,-200 40,-180 45,-160 48,-150 50,-144 60,-120 72,-100 75,-96 80,-90$$

Calculate the sum for each pair.

$$1-7200=-7199 2-3600=-3598 3-2400=-2397 4-1800=-1796 5-1440=-1435 6-1200=-1194 8-900=-892 9-800=-791 10-720=-710 12-600=-588 15-480=-465 16-450=-434 18-400=-382 20-360=-340 24-300=-276 25-288=-263 30-240=-210 32-225=-193 36-200=-164 40-180=-140 45-160=-115 48-150=-102 50-144=-94 60-120=-60 72-100=-28 75-96=-21 80-90=-10$$

The solution is the pair that gives sum -10.

$$a=-90 b=80$$

Rewrite x^{2}-10x-7200 as \left(x^{2}-90x\right)+\left(80x-7200\right).

$$\left(x^{2}-90x\right)+\left(80x-7200\right)$$

Factor out x in the first and 80 in the second group.

$$x\left(x-90\right)+80\left(x-90\right)$$

Factor out common term x-90 by using distributive property.

$$\left(x-90\right)\left(x+80\right)$$

To find equation solutions, solve x-90=0 and x+80=0.

$$x=90 x=-80$$

Steps Using the Quadratic Formula

Divide both sides by -5. Zero divided by any non-zero number gives zero.

$$0=x^{2}-10x-7200$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-10x-7200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -7200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-7200\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-7200\right)}}{2}$$

Multiply -4 times -7200.

$$x=\frac{-\left(-10\right)±\sqrt{100+28800}}{2}$$

Add 100 to 28800.

$$x=\frac{-\left(-10\right)±\sqrt{28900}}{2}$$

Take the square root of 28900.

$$x=\frac{-\left(-10\right)±170}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±170}{2}$$

Now solve the equation x=\frac{10±170}{2} when ± is plus. Add 10 to 170.

$$x=\frac{180}{2}$$

Divide 180 by 2.

$$x=90$$

Now solve the equation x=\frac{10±170}{2} when ± is minus. Subtract 170 from 10.

$$x=-\frac{160}{2}$$

Divide -160 by 2.

$$x=-80$$

The equation is now solved.

$$x=90 x=-80$$

Steps for Completing the Square

Divide both sides by -5. Zero divided by any non-zero number gives zero.

$$0=x^{2}-10x-7200$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-10x-7200=0$$

Add 7200 to both sides. Anything plus zero gives itself.

$$x^{2}-10x=7200$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=7200+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=7200+25$$

Add 7200 to 25.

$$x^{2}-10x+25=7225$$

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}.

$$\left(x-5\right)^{2}=7225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{7225}$$

Simplify.

$$x-5=85 x-5=-85$$

Add 5 to both sides of the equation.

$$x=90 x=-80$$