Solve x^2-35x=-150

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

Solve for x

$$x=5$$

$$x=30$$

Steps Using Factoring

Add 150 to both sides.

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

To solve the equation, factor x^{2}-35x+150 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.

$$a+b=-35 ab=150$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 150.

$$-1,-150 -2,-75 -3,-50 -5,-30 -6,-25 -10,-15$$

Calculate the sum for each pair.

$$-1-150=-151 -2-75=-77 -3-50=-53 -5-30=-35 -6-25=-31 -10-15=-25$$

The solution is the pair that gives sum -35.

$$a=-30 b=-5$$

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

$$\left(x-30\right)\left(x-5\right)$$

To find equation solutions, solve x-30=0 and x-5=0.

$$x=30 x=5$$

Steps Using Factoring By Grouping

Add 150 to both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+150. To find a and b, set up a system to be solved.

$$a+b=-35 ab=1\times 150=150$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 150.

$$-1,-150 -2,-75 -3,-50 -5,-30 -6,-25 -10,-15$$

Calculate the sum for each pair.

$$-1-150=-151 -2-75=-77 -3-50=-53 -5-30=-35 -6-25=-31 -10-15=-25$$

The solution is the pair that gives sum -35.

$$a=-30 b=-5$$

Rewrite x^{2}-35x+150 as \left(x^{2}-30x\right)+\left(-5x+150\right).

$$\left(x^{2}-30x\right)+\left(-5x+150\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-30\right)-5\left(x-30\right)$$

Factor out common term x-30 by using distributive property.

$$\left(x-30\right)\left(x-5\right)$$

To find equation solutions, solve x-30=0 and x-5=0.

$$x=30 x=5$$

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.

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

Add 150 to both sides of the equation.

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

Subtracting -150 from itself leaves 0.

$$x^{2}-35x-\left(-150\right)=0$$

Subtract -150 from 0.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -35 for b, and 150 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -35.

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

Multiply -4 times 150.

$$x=\frac{-\left(-35\right)±\sqrt{1225-600}}{2}$$

Add 1225 to -600.

$$x=\frac{-\left(-35\right)±\sqrt{625}}{2}$$

Take the square root of 625.

$$x=\frac{-\left(-35\right)±25}{2}$$

The opposite of -35 is 35.

$$x=\frac{35±25}{2}$$

Now solve the equation x=\frac{35±25}{2} when ± is plus. Add 35 to 25.

$$x=\frac{60}{2}$$

Divide 60 by 2.

$$x=30$$

Now solve the equation x=\frac{35±25}{2} when ± is minus. Subtract 25 from 35.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

The equation is now solved.

$$x=30 x=5$$

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.

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

Divide -35, the coefficient of the x term, by 2 to get -\frac{35}{2}. Then add the square of -\frac{35}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{35}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-35x+\frac{1225}{4}=-150+\frac{1225}{4}$$

Add -150 to \frac{1225}{4}.

$$x^{2}-35x+\frac{1225}{4}=\frac{625}{4}$$

Factor x^{2}-35x+\frac{1225}{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}.

$$\left(x-\frac{35}{2}\right)^{2}=\frac{625}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{2}\right)^{2}}=\sqrt{\frac{625}{4}}$$

Simplify.

$$x-\frac{35}{2}=\frac{25}{2} x-\frac{35}{2}=-\frac{25}{2}$$

Add \frac{35}{2} to both sides of the equation.

$$x=30 x=5$$