Solve r^2+3r=10

$$r ^ { 2 } + 3 r = 10$$

Solve for r

$$r=-5$$

$$r=2$$

Steps Using Factoring

Subtract 10 from both sides.

$$r^{2}+3r-10=0$$

To solve the equation, factor r^{2}+3r-10 using formula r^{2}+\left(a+b\right)r+ab=\left(r+a\right)\left(r+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

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

$$\left(r-2\right)\left(r+5\right)$$

To find equation solutions, solve r-2=0 and r+5=0.

$$r=2 r=-5$$

Steps Using Factoring By Grouping

Subtract 10 from both sides.

$$r^{2}+3r-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as r^{2}+ar+br-10. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-10\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

Rewrite r^{2}+3r-10 as \left(r^{2}-2r\right)+\left(5r-10\right).

$$\left(r^{2}-2r\right)+\left(5r-10\right)$$

Factor out r in the first and 5 in the second group.

$$r\left(r-2\right)+5\left(r-2\right)$$

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

$$\left(r-2\right)\left(r+5\right)$$

To find equation solutions, solve r-2=0 and r+5=0.

$$r=2 r=-5$$

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.

$$r^{2}+3r=10$$

Subtract 10 from both sides of the equation.

$$r^{2}+3r-10=10-10$$

Subtracting 10 from itself leaves 0.

$$r^{2}+3r-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-3±\sqrt{3^{2}-4\left(-10\right)}}{2}$$

Square 3.

$$r=\frac{-3±\sqrt{9-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$r=\frac{-3±\sqrt{9+40}}{2}$$

Add 9 to 40.

$$r=\frac{-3±\sqrt{49}}{2}$$

Take the square root of 49.

$$r=\frac{-3±7}{2}$$

Now solve the equation r=\frac{-3±7}{2} when ± is plus. Add -3 to 7.

$$r=\frac{4}{2}$$

Divide 4 by 2.

$$r=2$$

Now solve the equation r=\frac{-3±7}{2} when ± is minus. Subtract 7 from -3.

$$r=-\frac{10}{2}$$

Divide -10 by 2.

$$r=-5$$

The equation is now solved.

$$r=2 r=-5$$

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.

$$r^{2}+3r=10$$

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.

$$r^{2}+3r+\left(\frac{3}{2}\right)^{2}=10+\left(\frac{3}{2}\right)^{2}$$

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

$$r^{2}+3r+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$r^{2}+3r+\frac{9}{4}=\frac{49}{4}$$

Factor r^{2}+3r+\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}.

$$\left(r+\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$r+\frac{3}{2}=\frac{7}{2} r+\frac{3}{2}=-\frac{7}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$r=2 r=-5$$