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2m^2-40m+18=0
a = 2; b = -40; c = +18;
Δ = b2-4ac
Δ = -402-4·2·18
Δ = 1456
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1456}=\sqrt{16*91}=\sqrt{16}*\sqrt{91}=4\sqrt{91}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{91}}{2*2}=\frac{40-4\sqrt{91}}{4} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{91}}{2*2}=\frac{40+4\sqrt{91}}{4} $
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