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5x^2+32x+20=0
a = 5; b = 32; c = +20;
Δ = b2-4ac
Δ = 322-4·5·20
Δ = 624
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{39}}{2*5}=\frac{-32-4\sqrt{39}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{39}}{2*5}=\frac{-32+4\sqrt{39}}{10} $
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