(3m+4)(2m-5)=0

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Solution for (3m+4)(2m-5)=0 equation:



(3m+4)(2m-5)=0
We multiply parentheses ..
(+6m^2-15m+8m-20)=0
We get rid of parentheses
6m^2-15m+8m-20=0
We add all the numbers together, and all the variables
6m^2-7m-20=0
a = 6; b = -7; c = -20;
Δ = b2-4ac
Δ = -72-4·6·(-20)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-23}{2*6}=\frac{-16}{12} =-1+1/3 $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+23}{2*6}=\frac{30}{12} =2+1/2 $

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