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10m^2-53m+15=0
a = 10; b = -53; c = +15;
Δ = b2-4ac
Δ = -532-4·10·15
Δ = 2209
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{2209}=47$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-47}{2*10}=\frac{6}{20} =3/10 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+47}{2*10}=\frac{100}{20} =5 $
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