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3x(2x-5)=x+2
We move all terms to the left:
3x(2x-5)-(x+2)=0
We multiply parentheses
6x^2-15x-(x+2)=0
We get rid of parentheses
6x^2-15x-x-2=0
We add all the numbers together, and all the variables
6x^2-16x-2=0
a = 6; b = -16; c = -2;
Δ = b2-4ac
Δ = -162-4·6·(-2)
Δ = 304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{19}}{2*6}=\frac{16-4\sqrt{19}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{19}}{2*6}=\frac{16+4\sqrt{19}}{12} $
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