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2x^2=-4(2x+1)
We move all terms to the left:
2x^2-(-4(2x+1))=0
We calculate terms in parentheses: -(-4(2x+1)), so:We get rid of parentheses
-4(2x+1)
We multiply parentheses
-8x-4
Back to the equation:
-(-8x-4)
2x^2+8x+4=0
a = 2; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·2·4
Δ = 32
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{2}}{2*2}=\frac{-8-4\sqrt{2}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{2}}{2*2}=\frac{-8+4\sqrt{2}}{4} $
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