Solve -2x^2-8x+10/3=0

$$\frac{ -2 { x }^{ 2 } -8x+10 }{ 3 } =0$$

Solve for x

$$x=-5$$

$$x=1$$

Steps Using Factoring By Grouping

Multiply both sides by 3. Anything times zero gives zero.

$$-2x^{2}-8x+10=0$$

Divide both sides by 2.

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

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

$$a+b=-4 ab=-5=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-5$$

Rewrite -x^{2}-4x+5 as \left(-x^{2}+x\right)+\left(-5x+5\right).

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

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

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

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

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

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

$$x=1 x=-5$$

Steps Using the Quadratic Formula

Multiply both sides by 3. Anything times zero gives zero.

$$-2x^{2}-8x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -8 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-2\right)\times 10}}{2\left(-2\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-2\right)\times 10}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-8\right)±\sqrt{64+8\times 10}}{2\left(-2\right)}$$

Multiply 8 times 10.

$$x=\frac{-\left(-8\right)±\sqrt{64+80}}{2\left(-2\right)}$$

Add 64 to 80.

$$x=\frac{-\left(-8\right)±\sqrt{144}}{2\left(-2\right)}$$

Take the square root of 144.

$$x=\frac{-\left(-8\right)±12}{2\left(-2\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±12}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{8±12}{-4}$$

Now solve the equation x=\frac{8±12}{-4} when ± is plus. Add 8 to 12.

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

Divide 20 by -4.

$$x=-5$$

Now solve the equation x=\frac{8±12}{-4} when ± is minus. Subtract 12 from 8.

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

Divide -4 by -4.

$$x=1$$

The equation is now solved.

$$x=-5 x=1$$

Steps for Completing the Square

Multiply both sides by 3. Anything times zero gives zero.

$$-2x^{2}-8x+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}-8x=-10$$

Divide both sides by -2.

$$\frac{-2x^{2}-8x}{-2}=-\frac{10}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{8}{-2}\right)x=-\frac{10}{-2}$$

Divide -8 by -2.

$$x^{2}+4x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x^{2}+4x=5$$

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

$$x^{2}+4x+2^{2}=5+2^{2}$$

Square 2.

$$x^{2}+4x+4=5+4$$

Add 5 to 4.

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

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+2=3 x+2=-3$$

Subtract 2 from both sides of the equation.

$$x=1 x=-5$$