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10x^2+16x+4=0
a = 10; b = 16; c = +4;
Δ = b2-4ac
Δ = 162-4·10·4
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{6}}{2*10}=\frac{-16-4\sqrt{6}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{6}}{2*10}=\frac{-16+4\sqrt{6}}{20} $
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