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7m^2+45m+18=0
a = 7; b = 45; c = +18;
Δ = b2-4ac
Δ = 452-4·7·18
Δ = 1521
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{1521}=39$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-39}{2*7}=\frac{-84}{14} =-6 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+39}{2*7}=\frac{-6}{14} =-3/7 $
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