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2x^2+8x+3=0
a = 2; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·2·3
Δ = 40
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{10}}{2*2}=\frac{-8-2\sqrt{10}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{10}}{2*2}=\frac{-8+2\sqrt{10}}{4} $
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