Solve 3x^2+17x=-10

$$3 { x }^{ 2 } +17x=-10$$

Solve for x

$$x=-5$$

$$x=-\frac{2}{3}\approx -0.666666667$$

Steps Using Factoring By Grouping

Add 10 to both sides.

$$3x^{2}+17x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=17 ab=3\times 10=30$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.

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

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 17.

$$a=2 b=15$$

Rewrite 3x^{2}+17x+10 as \left(3x^{2}+2x\right)+\left(15x+10\right).

$$\left(3x^{2}+2x\right)+\left(15x+10\right)$$

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

$$x\left(3x+2\right)+5\left(3x+2\right)$$

Factor out common term 3x+2 by using distributive property.

$$\left(3x+2\right)\left(x+5\right)$$

To find equation solutions, solve 3x+2=0 and x+5=0.

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

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.

$$3x^{2}+17x=-10$$

Add 10 to both sides of the equation.

$$3x^{2}+17x-\left(-10\right)=-10-\left(-10\right)$$

Subtracting -10 from itself leaves 0.

$$3x^{2}+17x-\left(-10\right)=0$$

Subtract -10 from 0.

$$3x^{2}+17x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 17 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4\times 3\times 10}}{2\times 3}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 3\times 10}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-17±\sqrt{289-12\times 10}}{2\times 3}$$

Multiply -12 times 10.

$$x=\frac{-17±\sqrt{289-120}}{2\times 3}$$

Add 289 to -120.

$$x=\frac{-17±\sqrt{169}}{2\times 3}$$

Take the square root of 169.

$$x=\frac{-17±13}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-17±13}{6}$$

Now solve the equation x=\frac{-17±13}{6} when ± is plus. Add -17 to 13.

$$x=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-17±13}{6} when ± is minus. Subtract 13 from -17.

$$x=-\frac{30}{6}$$

Divide -30 by 6.

$$x=-5$$

The equation is now solved.

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

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.

$$3x^{2}+17x=-10$$

Divide both sides by 3.

$$\frac{3x^{2}+17x}{3}=-\frac{10}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{17}{3}x=-\frac{10}{3}$$

Divide \frac{17}{3}, the coefficient of the x term, by 2 to get \frac{17}{6}. Then add the square of \frac{17}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{17}{3}x+\left(\frac{17}{6}\right)^{2}=-\frac{10}{3}+\left(\frac{17}{6}\right)^{2}$$

Square \frac{17}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{17}{3}x+\frac{289}{36}=-\frac{10}{3}+\frac{289}{36}$$

Add -\frac{10}{3} to \frac{289}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{17}{3}x+\frac{289}{36}=\frac{169}{36}$$

Factor x^{2}+\frac{17}{3}x+\frac{289}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{17}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$x+\frac{17}{6}=\frac{13}{6} x+\frac{17}{6}=-\frac{13}{6}$$

Subtract \frac{17}{6} from both sides of the equation.

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