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7m^2+2m-3=0
a = 7; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·7·(-3)
Δ = 88
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{22}}{2*7}=\frac{-2-2\sqrt{22}}{14} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{22}}{2*7}=\frac{-2+2\sqrt{22}}{14} $
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