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4m^2-20m+16=0
a = 4; b = -20; c = +16;
Δ = b2-4ac
Δ = -202-4·4·16
Δ = 144
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{144}=12$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12}{2*4}=\frac{8}{8} =1 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12}{2*4}=\frac{32}{8} =4 $
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