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2m^2+10m+11=0
a = 2; b = 10; c = +11;
Δ = b2-4ac
Δ = 102-4·2·11
Δ = 12
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{3}}{2*2}=\frac{-10-2\sqrt{3}}{4} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{3}}{2*2}=\frac{-10+2\sqrt{3}}{4} $
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