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3x^2-5x-28=-6x^2-19x-20
We move all terms to the left:
3x^2-5x-28-(-6x^2-19x-20)=0
We get rid of parentheses
3x^2+6x^2+19x-5x+20-28=0
We add all the numbers together, and all the variables
9x^2+14x-8=0
a = 9; b = 14; c = -8;
Δ = b2-4ac
Δ = 142-4·9·(-8)
Δ = 484
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-22}{2*9}=\frac{-36}{18} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+22}{2*9}=\frac{8}{18} =4/9 $
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