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2x^2+16x-11=-7
We move all terms to the left:
2x^2+16x-11-(-7)=0
We add all the numbers together, and all the variables
2x^2+16x-4=0
a = 2; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·2·(-4)
Δ = 288
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-12\sqrt{2}}{2*2}=\frac{-16-12\sqrt{2}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+12\sqrt{2}}{2*2}=\frac{-16+12\sqrt{2}}{4} $
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