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(6x^2+36x+13)-(4x^2+13x+33)=0
We get rid of parentheses
6x^2-4x^2+36x-13x+13-33=0
We add all the numbers together, and all the variables
2x^2+23x-20=0
a = 2; b = 23; c = -20;
Δ = b2-4ac
Δ = 232-4·2·(-20)
Δ = 689
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{-(23)-\sqrt{689}}{2*2}=\frac{-23-\sqrt{689}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{689}}{2*2}=\frac{-23+\sqrt{689}}{4} $
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