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b^2+4b-10=0
a = 1; b = 4; c = -10;
Δ = b2-4ac
Δ = 42-4·1·(-10)
Δ = 56
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{14}}{2*1}=\frac{-4-2\sqrt{14}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{14}}{2*1}=\frac{-4+2\sqrt{14}}{2} $
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