Solve 75x^2+25x=2

$$75 { x }^{ 2 } +25x=2$$

Solve for x

$$x=-\frac{2}{5}=-0.4$$

$$x=\frac{1}{15}\approx 0.066666667$$

Steps Using Factoring By Grouping

Subtract 2 from both sides.

$$75x^{2}+25x-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 75x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=25 ab=75\left(-2\right)=-150$$

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 -150.

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

Calculate the sum for each pair.

$$-1+150=149 -2+75=73 -3+50=47 -5+30=25 -6+25=19 -10+15=5$$

The solution is the pair that gives sum 25.

$$a=-5 b=30$$

Rewrite 75x^{2}+25x-2 as \left(75x^{2}-5x\right)+\left(30x-2\right).

$$\left(75x^{2}-5x\right)+\left(30x-2\right)$$

Factor out 5x in the first and 2 in the second group.

$$5x\left(15x-1\right)+2\left(15x-1\right)$$

Factor out common term 15x-1 by using distributive property.

$$\left(15x-1\right)\left(5x+2\right)$$

To find equation solutions, solve 15x-1=0 and 5x+2=0.

$$x=\frac{1}{15} x=-\frac{2}{5}$$

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.

$$75x^{2}+25x=2$$

Subtract 2 from both sides of the equation.

$$75x^{2}+25x-2=2-2$$

Subtracting 2 from itself leaves 0.

$$75x^{2}+25x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 75 for a, 25 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\times 75\left(-2\right)}}{2\times 75}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 75\left(-2\right)}}{2\times 75}$$

Multiply -4 times 75.

$$x=\frac{-25±\sqrt{625-300\left(-2\right)}}{2\times 75}$$

Multiply -300 times -2.

$$x=\frac{-25±\sqrt{625+600}}{2\times 75}$$

Add 625 to 600.

$$x=\frac{-25±\sqrt{1225}}{2\times 75}$$

Take the square root of 1225.

$$x=\frac{-25±35}{2\times 75}$$

Multiply 2 times 75.

$$x=\frac{-25±35}{150}$$

Now solve the equation x=\frac{-25±35}{150} when ± is plus. Add -25 to 35.

$$x=\frac{10}{150}$$

Reduce the fraction \frac{10}{150} to lowest terms by extracting and canceling out 10.

$$x=\frac{1}{15}$$

Now solve the equation x=\frac{-25±35}{150} when ± is minus. Subtract 35 from -25.

$$x=-\frac{60}{150}$$

Reduce the fraction \frac{-60}{150} to lowest terms by extracting and canceling out 30.

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

The equation is now solved.

$$x=\frac{1}{15} x=-\frac{2}{5}$$

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.

$$75x^{2}+25x=2$$

Divide both sides by 75.

$$\frac{75x^{2}+25x}{75}=\frac{2}{75}$$

Dividing by 75 undoes the multiplication by 75.

$$x^{2}+\frac{25}{75}x=\frac{2}{75}$$

Reduce the fraction \frac{25}{75} to lowest terms by extracting and canceling out 25.

$$x^{2}+\frac{1}{3}x=\frac{2}{75}$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{2}{75}+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{2}{75}+\frac{1}{36}$$

Add \frac{2}{75} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{49}{900}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{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}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{49}{900}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{49}{900}}$$

Simplify.

$$x+\frac{1}{6}=\frac{7}{30} x+\frac{1}{6}=-\frac{7}{30}$$

Subtract \frac{1}{6} from both sides of the equation.

$$x=\frac{1}{15} x=-\frac{2}{5}$$