Solve 5x^2+-35x+-150=0

$$5 x ^ { 2 } + - 35 x + - 150 = 0$$

Solve for x

$$x=-3$$

$$x=10$$

Steps Using Factoring By Grouping

Divide both sides by 5.

$$x^{2}-7x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\left(-30\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -7.

$$a=-10 b=3$$

Rewrite x^{2}-7x-30 as \left(x^{2}-10x\right)+\left(3x-30\right).

$$\left(x^{2}-10x\right)+\left(3x-30\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-10\right)+3\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+3\right)$$

To find equation solutions, solve x-10=0 and x+3=0.

$$x=10 x=-3$$

Steps Using the Quadratic Formula

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.

$$5x^{2}-35x-150=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -35 for b, and -150 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 5\left(-150\right)}}{2\times 5}$$

Square -35.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 5\left(-150\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-35\right)±\sqrt{1225-20\left(-150\right)}}{2\times 5}$$

Multiply -20 times -150.

$$x=\frac{-\left(-35\right)±\sqrt{1225+3000}}{2\times 5}$$

Add 1225 to 3000.

$$x=\frac{-\left(-35\right)±\sqrt{4225}}{2\times 5}$$

Take the square root of 4225.

$$x=\frac{-\left(-35\right)±65}{2\times 5}$$

The opposite of -35 is 35.

$$x=\frac{35±65}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{35±65}{10}$$

Now solve the equation x=\frac{35±65}{10} when ± is plus. Add 35 to 65.

$$x=\frac{100}{10}$$

Divide 100 by 10.

$$x=10$$

Now solve the equation x=\frac{35±65}{10} when ± is minus. Subtract 65 from 35.

$$x=-\frac{30}{10}$$

Divide -30 by 10.

$$x=-3$$

The equation is now solved.

$$x=10 x=-3$$

Steps for Completing the Square

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.

$$5x^{2}-35x-150=0$$

Add 150 to both sides of the equation.

$$5x^{2}-35x-150-\left(-150\right)=-\left(-150\right)$$

Subtracting -150 from itself leaves 0.

$$5x^{2}-35x=-\left(-150\right)$$

Subtract -150 from 0.

$$5x^{2}-35x=150$$

Divide both sides by 5.

$$\frac{5x^{2}-35x}{5}=\frac{150}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{35}{5}\right)x=\frac{150}{5}$$

Divide -35 by 5.

$$x^{2}-7x=\frac{150}{5}$$

Divide 150 by 5.

$$x^{2}-7x=30$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=30+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=30+\frac{49}{4}$$

Add 30 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{169}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{13}{2} x-\frac{7}{2}=-\frac{13}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=10 x=-3$$