(m-3)(m-7)=

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Solution for (m-3)(m-7)= equation:



(m-3)(m-7)=
We move all terms to the left:
(m-3)(m-7)-()=0
We add all the numbers together, and all the variables
(m-3)(m-7)=0
We multiply parentheses ..
(+m^2-7m-3m+21)=0
We get rid of parentheses
m^2-7m-3m+21=0
We add all the numbers together, and all the variables
m^2-10m+21=0
a = 1; b = -10; c = +21;
Δ = b2-4ac
Δ = -102-4·1·21
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4}{2*1}=\frac{6}{2} =3 $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4}{2*1}=\frac{14}{2} =7 $

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