Solve 2x+2=5x^2-5

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

Solve for x

$$x=-1$$

$$x = \frac{7}{5} = 1\frac{2}{5} = 1.4$$

Steps Using Factoring By Grouping

Subtract 5x^{2} from both sides.

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

Add 5 to both sides.

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

Add 2 and 5 to get 7.

$$2x+7-5x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-5x^{2}+2x+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=2 ab=-5\times 7=-35$$

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 -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$a=7 b=-5$$

Rewrite -5x^{2}+2x+7 as \left(-5x^{2}+7x\right)+\left(-5x+7\right).

$$\left(-5x^{2}+7x\right)+\left(-5x+7\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(5x-7\right)-\left(5x-7\right)$$

Factor out common term 5x-7 by using distributive property.

$$\left(5x-7\right)\left(-x-1\right)$$

To find equation solutions, solve 5x-7=0 and -x-1=0.

$$x=\frac{7}{5} x=-1$$

Steps Using the Quadratic Formula

Subtract 5x^{2} from both sides.

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

Add 5 to both sides.

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

Add 2 and 5 to get 7.

$$2x+7-5x^{2}=0$$

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.

$$-5x^{2}+2x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 2 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-5\right)\times 7}}{2\left(-5\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-5\right)\times 7}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-2±\sqrt{4+20\times 7}}{2\left(-5\right)}$$

Multiply 20 times 7.

$$x=\frac{-2±\sqrt{4+140}}{2\left(-5\right)}$$

Add 4 to 140.

$$x=\frac{-2±\sqrt{144}}{2\left(-5\right)}$$

Take the square root of 144.

$$x=\frac{-2±12}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-2±12}{-10}$$

Now solve the equation x=\frac{-2±12}{-10} when ± is plus. Add -2 to 12.

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

Divide 10 by -10.

$$x=-1$$

Now solve the equation x=\frac{-2±12}{-10} when ± is minus. Subtract 12 from -2.

$$x=-\frac{14}{-10}$$

Reduce the fraction \frac{-14}{-10} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{5}$$

The equation is now solved.

$$x=-1 x=\frac{7}{5}$$

Steps for Completing the Square

Subtract 5x^{2} from both sides.

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

Subtract 2 from both sides.

$$2x-5x^{2}=-5-2$$

Subtract 2 from -5 to get -7.

$$2x-5x^{2}=-7$$

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.

$$-5x^{2}+2x=-7$$

Divide both sides by -5.

$$\frac{-5x^{2}+2x}{-5}=-\frac{7}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{2}{-5}x=-\frac{7}{-5}$$

Divide 2 by -5.

$$x^{2}-\frac{2}{5}x=-\frac{7}{-5}$$

Divide -7 by -5.

$$x^{2}-\frac{2}{5}x=\frac{7}{5}$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=\frac{7}{5}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{7}{5}+\frac{1}{25}$$

Add \frac{7}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{36}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{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}.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{6}{5} x-\frac{1}{5}=-\frac{6}{5}$$

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{7}{5} x=-1$$