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4x^2+20x+14=0
a = 4; b = 20; c = +14;
Δ = b2-4ac
Δ = 202-4·4·14
Δ = 176
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{11}}{2*4}=\frac{-20-4\sqrt{11}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{11}}{2*4}=\frac{-20+4\sqrt{11}}{8} $
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