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3m(m+5)=36
We move all terms to the left:
3m(m+5)-(36)=0
We multiply parentheses
3m^2+15m-36=0
a = 3; b = 15; c = -36;
Δ = b2-4ac
Δ = 152-4·3·(-36)
Δ = 657
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{73}}{2*3}=\frac{-15-3\sqrt{73}}{6} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{73}}{2*3}=\frac{-15+3\sqrt{73}}{6} $
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