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2x^2+3x^2-2x-3=0
We add all the numbers together, and all the variables
5x^2-2x-3=0
a = 5; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·5·(-3)
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*5}=\frac{-6}{10} =-3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*5}=\frac{10}{10} =1 $
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