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2x-(2x^2-3x-10)=-2
We move all terms to the left:
2x-(2x^2-3x-10)-(-2)=0
We add all the numbers together, and all the variables
2x-(2x^2-3x-10)+2=0
We get rid of parentheses
-2x^2+2x+3x+10+2=0
We add all the numbers together, and all the variables
-2x^2+5x+12=0
a = -2; b = 5; c = +12;
Δ = b2-4ac
Δ = 52-4·(-2)·12
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*-2}=\frac{-16}{-4} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*-2}=\frac{6}{-4} =-1+1/2 $
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