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10x^2+4x-5=0
a = 10; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·10·(-5)
Δ = 216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{6}}{2*10}=\frac{-4-6\sqrt{6}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{6}}{2*10}=\frac{-4+6\sqrt{6}}{20} $
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