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-16=2x^2+2x-40
We move all terms to the left:
-16-(2x^2+2x-40)=0
We get rid of parentheses
-2x^2-2x+40-16=0
We add all the numbers together, and all the variables
-2x^2-2x+24=0
a = -2; b = -2; c = +24;
Δ = b2-4ac
Δ = -22-4·(-2)·24
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*-2}=\frac{-12}{-4} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*-2}=\frac{16}{-4} =-4 $
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