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3x+x(x+4)=79
We move all terms to the left:
3x+x(x+4)-(79)=0
We multiply parentheses
x^2+3x+4x-79=0
We add all the numbers together, and all the variables
x^2+7x-79=0
a = 1; b = 7; c = -79;
Δ = b2-4ac
Δ = 72-4·1·(-79)
Δ = 365
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{365}}{2*1}=\frac{-7-\sqrt{365}}{2}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{365}}{2*1}=\frac{-7+\sqrt{365}}{2}
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