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10x^2-20x-12=-6x
We move all terms to the left:
10x^2-20x-12-(-6x)=0
We get rid of parentheses
10x^2-20x+6x-12=0
We add all the numbers together, and all the variables
10x^2-14x-12=0
a = 10; b = -14; c = -12;
Δ = b2-4ac
Δ = -142-4·10·(-12)
Δ = 676
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{676}=26$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-26}{2*10}=\frac{-12}{20} =-3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+26}{2*10}=\frac{40}{20} =2 $
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