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3m^2+10m-32=0
a = 3; b = 10; c = -32;
Δ = b2-4ac
Δ = 102-4·3·(-32)
Δ = 484
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*3}=\frac{-32}{6} =-5+1/3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*3}=\frac{12}{6} =2 $
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