r^2+12-7r=0

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Solution for r^2+12-7r=0 equation: 



r^2+12-7r=0

a = 1; b = -7; c = +12;
Δ = b2-4ac
Δ = -72-4·1·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*1}=\frac{6}{2}
=3
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*1}=\frac{8}{2}
=4
$

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