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(s-3)(s-4)=6
We move all terms to the left:
(s-3)(s-4)-(6)=0
We multiply parentheses ..
(+s^2-4s-3s+12)-6=0
We get rid of parentheses
s^2-4s-3s+12-6=0
We add all the numbers together, and all the variables
s^2-7s+6=0
a = 1; b = -7; c = +6;
Δ = b2-4ac
Δ = -72-4·1·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-5}{2*1}=\frac{2}{2} =1 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+5}{2*1}=\frac{12}{2} =6 $
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