Solve 2x^2+7x-1=4-2x

$$2 x ^ { 2 } + 7 x - 1 = 4 - 2 x$$

Solve for x

$$x=-5$$

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Subtract 4 from both sides.

$$2x^{2}+7x-1-4=-2x$$

Subtract 4 from -1 to get -5.

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

Add 2x to both sides.

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

Combine 7x and 2x to get 9x.

$$2x^{2}+9x-5=0$$

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

$$a+b=9 ab=2\left(-5\right)=-10$$

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 9.

$$a=-1 b=10$$

Rewrite 2x^{2}+9x-5 as \left(2x^{2}-x\right)+\left(10x-5\right).

$$\left(2x^{2}-x\right)+\left(10x-5\right)$$

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

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

Factor out common term 2x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$2x^{2}+7x-1-4=-2x$$

Subtract 4 from -1 to get -5.

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

Add 2x to both sides.

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

Combine 7x and 2x to get 9x.

$$2x^{2}+9x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 9 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-9±\sqrt{81-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-9±\sqrt{81+40}}{2\times 2}$$

Add 81 to 40.

$$x=\frac{-9±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-9±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-9±11}{4}$$

Now solve the equation x=\frac{-9±11}{4} when ± is plus. Add -9 to 11.

$$x=\frac{2}{4}$$

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

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

Now solve the equation x=\frac{-9±11}{4} when ± is minus. Subtract 11 from -9.

$$x=-\frac{20}{4}$$

Divide -20 by 4.

$$x=-5$$

The equation is now solved.

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

Steps for Completing the Square

Add 2x to both sides.

$$2x^{2}+7x-1+2x=4$$

Combine 7x and 2x to get 9x.

$$2x^{2}+9x-1=4$$

Add 1 to both sides.

$$2x^{2}+9x=4+1$$

Add 4 and 1 to get 5.

$$2x^{2}+9x=5$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{2}x+\left(\frac{9}{4}\right)^{2}=\frac{5}{2}+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{5}{2}+\frac{81}{16}$$

Add \frac{5}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{121}{16}$$

Factor x^{2}+\frac{9}{2}x+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x+\frac{9}{4}=\frac{11}{4} x+\frac{9}{4}=-\frac{11}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

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