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8x+4x^2=240
We move all terms to the left:
8x+4x^2-(240)=0
a = 4; b = 8; c = -240;
Δ = b2-4ac
Δ = 82-4·4·(-240)
Δ = 3904
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3904}=\sqrt{64*61}=\sqrt{64}*\sqrt{61}=8\sqrt{61}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{61}}{2*4}=\frac{-8-8\sqrt{61}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{61}}{2*4}=\frac{-8+8\sqrt{61}}{8} $
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