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(9x^2+6x+5)+(3x^2-5x-9)=0
We get rid of parentheses
9x^2+3x^2+6x-5x+5-9=0
We add all the numbers together, and all the variables
12x^2+x-4=0
a = 12; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·12·(-4)
Δ = 193
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{-(1)-\sqrt{193}}{2*12}=\frac{-1-\sqrt{193}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{193}}{2*12}=\frac{-1+\sqrt{193}}{24} $
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