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4x^2+14x-8=0
a = 4; b = 14; c = -8;
Δ = b2-4ac
Δ = 142-4·4·(-8)
Δ = 324
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-18}{2*4}=\frac{-32}{8} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+18}{2*4}=\frac{4}{8} =1/2 $
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