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-4x+6x^2=10
We move all terms to the left:
-4x+6x^2-(10)=0
a = 6; b = -4; c = -10;
Δ = b2-4ac
Δ = -42-4·6·(-10)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*6}=\frac{-12}{12} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*6}=\frac{20}{12} =1+2/3 $
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