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(x+3)(2x+11)=0
We multiply parentheses ..
(+2x^2+11x+6x+33)=0
We get rid of parentheses
2x^2+11x+6x+33=0
We add all the numbers together, and all the variables
2x^2+17x+33=0
a = 2; b = 17; c = +33;
Δ = b2-4ac
Δ = 172-4·2·33
Δ = 25
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-5}{2*2}=\frac{-22}{4} =-5+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+5}{2*2}=\frac{-12}{4} =-3 $
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