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2x^2+6=-8x
We move all terms to the left:
2x^2+6-(-8x)=0
We get rid of parentheses
2x^2+8x+6=0
a = 2; b = 8; c = +6;
Δ = b2-4ac
Δ = 82-4·2·6
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*2}=\frac{-12}{4} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*2}=\frac{-4}{4} =-1 $
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