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21m^2-52m+32=0
a = 21; b = -52; c = +32;
Δ = b2-4ac
Δ = -522-4·21·32
Δ = 16
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{16}=4$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-4}{2*21}=\frac{48}{42} =1+1/7 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+4}{2*21}=\frac{56}{42} =1+1/3 $
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