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(x+3)(x+4)=35
We move all terms to the left:
(x+3)(x+4)-(35)=0
We multiply parentheses ..
(+x^2+4x+3x+12)-35=0
We get rid of parentheses
x^2+4x+3x+12-35=0
We add all the numbers together, and all the variables
x^2+7x-23=0
a = 1; b = 7; c = -23;
Δ = b2-4ac
Δ = 72-4·1·(-23)
Δ = 141
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{141}}{2*1}=\frac{-7-\sqrt{141}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{141}}{2*1}=\frac{-7+\sqrt{141}}{2} $
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