Solve -2x^2-8x+2+19x-7=0

$$- 2 x ^ { 2 } - 8 x + 2 + 19 x - 7 = 0$$

Solve for x

$$x=5$$

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

Steps Using Factoring By Grouping

Combine -8x and 19x to get 11x.

$$-2x^{2}+11x+2-7=0$$

Subtract 7 from 2 to get -5.

$$-2x^{2}+11x-5=0$$

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.

$$a+b=11 ab=-2\left(-5\right)=10$$

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 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 11.

$$a=10 b=1$$

Rewrite -2x^{2}+11x-5 as \left(-2x^{2}+10x\right)+\left(x-5\right).

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

Factor out 2x in the first and -1 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Combine -8x and 19x to get 11x.

$$-2x^{2}+11x+2-7=0$$

Subtract 7 from 2 to get -5.

$$-2x^{2}+11x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 11 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-2\right)\left(-5\right)}}{2\left(-2\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-2\right)\left(-5\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-11±\sqrt{121+8\left(-5\right)}}{2\left(-2\right)}$$

Multiply 8 times -5.

$$x=\frac{-11±\sqrt{121-40}}{2\left(-2\right)}$$

Add 121 to -40.

$$x=\frac{-11±\sqrt{81}}{2\left(-2\right)}$$

Take the square root of 81.

$$x=\frac{-11±9}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-11±9}{-4}$$

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

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

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{-11±9}{-4} when ± is minus. Subtract 9 from -11.

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

Divide -20 by -4.

$$x=5$$

The equation is now solved.

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

Steps for Completing the Square

Combine -8x and 19x to get 11x.

$$-2x^{2}+11x+2-7=0$$

Subtract 7 from 2 to get -5.

$$-2x^{2}+11x-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$-2x^{2}+11x=5$$

Divide both sides by -2.

$$\frac{-2x^{2}+11x}{-2}=\frac{5}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{11}{-2}x=\frac{5}{-2}$$

Divide 11 by -2.

$$x^{2}-\frac{11}{2}x=\frac{5}{-2}$$

Divide 5 by -2.

$$x^{2}-\frac{11}{2}x=-\frac{5}{2}$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-\frac{5}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-\frac{5}{2}+\frac{121}{16}$$

Add -\frac{5}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{81}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{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}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

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

Add \frac{11}{4} to both sides of the equation.

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