Solve 5x^2-15x=-10quad

$$5 x ^ { 2 } - 15 x = - 10 \quad$$

Solve for x

$$x=1$$

$$x=2$$

Steps Using Factoring By Grouping

Add 10 to both sides.

$$5x^{2}-15x+10=0$$

Divide both sides by 5.

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

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

$$a+b=-3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-2 b=-1$$

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

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

Factor out x in the first and -1 in the second group.

$$x\left(x-2\right)-\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x-1\right)$$

To find equation solutions, solve x-2=0 and x-1=0.

$$x=2 x=1$$

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.

$$5x^{2}-15x=-10$$

Add 10 to both sides of the equation.

$$5x^{2}-15x-\left(-10\right)=-10-\left(-10\right)$$

Subtracting -10 from itself leaves 0.

$$5x^{2}-15x-\left(-10\right)=0$$

Subtract -10 from 0.

$$5x^{2}-15x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -15 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 5\times 10}}{2\times 5}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 5\times 10}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-15\right)±\sqrt{225-20\times 10}}{2\times 5}$$

Multiply -20 times 10.

$$x=\frac{-\left(-15\right)±\sqrt{225-200}}{2\times 5}$$

Add 225 to -200.

$$x=\frac{-\left(-15\right)±\sqrt{25}}{2\times 5}$$

Take the square root of 25.

$$x=\frac{-\left(-15\right)±5}{2\times 5}$$

The opposite of -15 is 15.

$$x=\frac{15±5}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{15±5}{10}$$

Now solve the equation x=\frac{15±5}{10} when ± is plus. Add 15 to 5.

$$x=\frac{20}{10}$$

Divide 20 by 10.

$$x=2$$

Now solve the equation x=\frac{15±5}{10} when ± is minus. Subtract 5 from 15.

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

Divide 10 by 10.

$$x=1$$

The equation is now solved.

$$x=2 x=1$$

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.

$$5x^{2}-15x=-10$$

Divide both sides by 5.

$$\frac{5x^{2}-15x}{5}=-\frac{10}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{15}{5}\right)x=-\frac{10}{5}$$

Divide -15 by 5.

$$x^{2}-3x=-\frac{10}{5}$$

Divide -10 by 5.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{1}{2} x-\frac{3}{2}=-\frac{1}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$