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2m^2-32m+78=0
a = 2; b = -32; c = +78;
Δ = b2-4ac
Δ = -322-4·2·78
Δ = 400
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{400}=20$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-20}{2*2}=\frac{12}{4} =3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+20}{2*2}=\frac{52}{4} =13 $
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