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10x^2+6x-5=0
a = 10; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·10·(-5)
Δ = 236
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{236}=\sqrt{4*59}=\sqrt{4}*\sqrt{59}=2\sqrt{59}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{59}}{2*10}=\frac{-6-2\sqrt{59}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{59}}{2*10}=\frac{-6+2\sqrt{59}}{20} $
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