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4m^2-21m+10=0
a = 4; b = -21; c = +10;
Δ = b2-4ac
Δ = -212-4·4·10
Δ = 281
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{281}}{2*4}=\frac{21-\sqrt{281}}{8} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{281}}{2*4}=\frac{21+\sqrt{281}}{8} $
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