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(-13x^2-13x-10)+(19x^2-19x-5)=0
We get rid of parentheses
-13x^2+19x^2-13x-19x-10-5=0
We add all the numbers together, and all the variables
6x^2-32x-15=0
a = 6; b = -32; c = -15;
Δ = b2-4ac
Δ = -322-4·6·(-15)
Δ = 1384
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1384}=\sqrt{4*346}=\sqrt{4}*\sqrt{346}=2\sqrt{346}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{346}}{2*6}=\frac{32-2\sqrt{346}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{346}}{2*6}=\frac{32+2\sqrt{346}}{12} $
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