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3m+12=-m(m+13)
We move all terms to the left:
3m+12-(-m(m+13))=0
We calculate terms in parentheses: -(-m(m+13)), so:We get rid of parentheses
-m(m+13)
We multiply parentheses
-m^2-13m
We add all the numbers together, and all the variables
-1m^2-13m
Back to the equation:
-(-1m^2-13m)
1m^2+13m+3m+12=0
We add all the numbers together, and all the variables
m^2+16m+12=0
a = 1; b = 16; c = +12;
Δ = b2-4ac
Δ = 162-4·1·12
Δ = 208
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{13}}{2*1}=\frac{-16-4\sqrt{13}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{13}}{2*1}=\frac{-16+4\sqrt{13}}{2} $
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