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=-5x^2-3x+2
We move all terms to the left:
-(-5x^2-3x+2)=0
We get rid of parentheses
5x^2+3x-2=0
a = 5; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·5·(-2)
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*5}=\frac{-10}{10} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*5}=\frac{4}{10} =2/5 $
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