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10x^2-2x-5=4x^2-21x+2
We move all terms to the left:
10x^2-2x-5-(4x^2-21x+2)=0
We get rid of parentheses
10x^2-4x^2-2x+21x-2-5=0
We add all the numbers together, and all the variables
6x^2+19x-7=0
a = 6; b = 19; c = -7;
Δ = b2-4ac
Δ = 192-4·6·(-7)
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-23}{2*6}=\frac{-42}{12} =-3+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+23}{2*6}=\frac{4}{12} =1/3 $
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