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7x(3+x)=2x+5
We move all terms to the left:
7x(3+x)-(2x+5)=0
We add all the numbers together, and all the variables
7x(x+3)-(2x+5)=0
We multiply parentheses
7x^2+21x-(2x+5)=0
We get rid of parentheses
7x^2+21x-2x-5=0
We add all the numbers together, and all the variables
7x^2+19x-5=0
a = 7; b = 19; c = -5;
Δ = b2-4ac
Δ = 192-4·7·(-5)
Δ = 501
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{-(19)-\sqrt{501}}{2*7}=\frac{-19-\sqrt{501}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{501}}{2*7}=\frac{-19+\sqrt{501}}{14} $
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