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(6x-5)+(4x+5)x=2
We move all terms to the left:
(6x-5)+(4x+5)x-(2)=0
We multiply parentheses
4x^2+(6x-5)+5x-2=0
We get rid of parentheses
4x^2+6x+5x-5-2=0
We add all the numbers together, and all the variables
4x^2+11x-7=0
a = 4; b = 11; c = -7;
Δ = b2-4ac
Δ = 112-4·4·(-7)
Δ = 233
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{-(11)-\sqrt{233}}{2*4}=\frac{-11-\sqrt{233}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{233}}{2*4}=\frac{-11+\sqrt{233}}{8} $
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