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4x^2+8x-336=0
a = 4; b = 8; c = -336;
Δ = b2-4ac
Δ = 82-4·4·(-336)
Δ = 5440
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5440}=\sqrt{64*85}=\sqrt{64}*\sqrt{85}=8\sqrt{85}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{85}}{2*4}=\frac{-8-8\sqrt{85}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{85}}{2*4}=\frac{-8+8\sqrt{85}}{8} $
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