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(f+1)(f-6)=0
We multiply parentheses ..
(+f^2-6f+f-6)=0
We get rid of parentheses
f^2-6f+f-6=0
We add all the numbers together, and all the variables
f^2-5f-6=0
a = 1; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·1·(-6)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*1}=\frac{-2}{2} =-1 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*1}=\frac{12}{2} =6 $
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