Solve 5x^2-7x-6=-10x-4

$$5 x ^ { 2 } - 7 x - 6 = - 10 x - 4$$

Solve for x

$$x=-1$$

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

Steps Using Factoring By Grouping

Add 10x to both sides.

$$5x^{2}-7x-6+10x=-4$$

Combine -7x and 10x to get 3x.

$$5x^{2}+3x-6=-4$$

Add 4 to both sides.

$$5x^{2}+3x-6+4=0$$

Add -6 and 4 to get -2.

$$5x^{2}+3x-2=0$$

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

$$a+b=3 ab=5\left(-2\right)=-10$$

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 5x^{2}+3x-2 as \left(5x^{2}-2x\right)+\left(5x-2\right).

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

Factor out x in 5x^{2}-2x.

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

Factor out common term 5x-2 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Add 10x to both sides.

$$5x^{2}-7x-6+10x=-4$$

Combine -7x and 10x to get 3x.

$$5x^{2}+3x-6=-4$$

Add 4 to both sides.

$$5x^{2}+3x-6+4=0$$

Add -6 and 4 to get -2.

$$5x^{2}+3x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 5\left(-2\right)}}{2\times 5}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 5\left(-2\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-3±\sqrt{9-20\left(-2\right)}}{2\times 5}$$

Multiply -20 times -2.

$$x=\frac{-3±\sqrt{9+40}}{2\times 5}$$

Add 9 to 40.

$$x=\frac{-3±\sqrt{49}}{2\times 5}$$

Take the square root of 49.

$$x=\frac{-3±7}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-3±7}{10}$$

Now solve the equation x=\frac{-3±7}{10} when ± is plus. Add -3 to 7.

$$x=\frac{4}{10}$$

Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-3±7}{10} when ± is minus. Subtract 7 from -3.

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

Divide -10 by 10.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Add 10x to both sides.

$$5x^{2}-7x-6+10x=-4$$

Combine -7x and 10x to get 3x.

$$5x^{2}+3x-6=-4$$

Add 6 to both sides.

$$5x^{2}+3x=-4+6$$

Add -4 and 6 to get 2.

$$5x^{2}+3x=2$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide \frac{3}{5}, the coefficient of the x term, by 2 to get \frac{3}{10}. Then add the square of \frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=\frac{2}{5}+\left(\frac{3}{10}\right)^{2}$$

Square \frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{2}{5}+\frac{9}{100}$$

Add \frac{2}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{49}{100}$$

Factor x^{2}+\frac{3}{5}x+\frac{9}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{10}\right)^{2}=\frac{49}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{\frac{49}{100}}$$

Simplify.

$$x+\frac{3}{10}=\frac{7}{10} x+\frac{3}{10}=-\frac{7}{10}$$

Subtract \frac{3}{10} from both sides of the equation.

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