Solve 5x^2+28x+15=0

$$5 { x }^{ 2 } +28x+15=0$$

Solve for x

$$x=-5$$

$$x=-\frac{3}{5}=-0.6$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=28 ab=5\times 15=75$$

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 75.

$$1,75 3,25 5,15$$

Calculate the sum for each pair.

$$1+75=76 3+25=28 5+15=20$$

The solution is the pair that gives sum 28.

$$a=3 b=25$$

Rewrite 5x^{2}+28x+15 as \left(5x^{2}+3x\right)+\left(25x+15\right).

$$\left(5x^{2}+3x\right)+\left(25x+15\right)$$

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

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

Factor out common term 5x+3 by using distributive property.

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

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

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

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.

$$5x^{2}+28x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 28 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-28±\sqrt{28^{2}-4\times 5\times 15}}{2\times 5}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\times 5\times 15}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-28±\sqrt{784-20\times 15}}{2\times 5}$$

Multiply -20 times 15.

$$x=\frac{-28±\sqrt{784-300}}{2\times 5}$$

Add 784 to -300.

$$x=\frac{-28±\sqrt{484}}{2\times 5}$$

Take the square root of 484.

$$x=\frac{-28±22}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-28±22}{10}$$

Now solve the equation x=\frac{-28±22}{10} when ± is plus. Add -28 to 22.

$$x=-\frac{6}{10}$$

Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-28±22}{10} when ± is minus. Subtract 22 from -28.

$$x=-\frac{50}{10}$$

Divide -50 by 10.

$$x=-5$$

The equation is now solved.

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

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.

$$5x^{2}+28x+15=0$$

Subtract 15 from both sides of the equation.

$$5x^{2}+28x+15-15=-15$$

Subtracting 15 from itself leaves 0.

$$5x^{2}+28x=-15$$

Divide both sides by 5.

$$\frac{5x^{2}+28x}{5}=-\frac{15}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{28}{5}x=-\frac{15}{5}$$

Divide -15 by 5.

$$x^{2}+\frac{28}{5}x=-3$$

Divide \frac{28}{5}, the coefficient of the x term, by 2 to get \frac{14}{5}. Then add the square of \frac{14}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{28}{5}x+\left(\frac{14}{5}\right)^{2}=-3+\left(\frac{14}{5}\right)^{2}$$

Square \frac{14}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{28}{5}x+\frac{196}{25}=-3+\frac{196}{25}$$

Add -3 to \frac{196}{25}.

$$x^{2}+\frac{28}{5}x+\frac{196}{25}=\frac{121}{25}$$

Factor x^{2}+\frac{28}{5}x+\frac{196}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{14}{5}\right)^{2}=\frac{121}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{14}{5}\right)^{2}}=\sqrt{\frac{121}{25}}$$

Simplify.

$$x+\frac{14}{5}=\frac{11}{5} x+\frac{14}{5}=-\frac{11}{5}$$

Subtract \frac{14}{5} from both sides of the equation.

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