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4x^2+8x-390=0
a = 4; b = 8; c = -390;
Δ = b2-4ac
Δ = 82-4·4·(-390)
Δ = 6304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6304}=\sqrt{16*394}=\sqrt{16}*\sqrt{394}=4\sqrt{394}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{394}}{2*4}=\frac{-8-4\sqrt{394}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{394}}{2*4}=\frac{-8+4\sqrt{394}}{8} $
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