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4m^2+10m-5=0
a = 4; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·4·(-5)
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{5}}{2*4}=\frac{-10-6\sqrt{5}}{8} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{5}}{2*4}=\frac{-10+6\sqrt{5}}{8} $
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