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5b^2-4b-10=0
a = 5; b = -4; c = -10;
Δ = b2-4ac
Δ = -42-4·5·(-10)
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{6}}{2*5}=\frac{4-6\sqrt{6}}{10} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{6}}{2*5}=\frac{4+6\sqrt{6}}{10} $
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