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-3=4x^2-1x-3
We move all terms to the left:
-3-(4x^2-1x-3)=0
We get rid of parentheses
-4x^2+1x+3-3=0
We add all the numbers together, and all the variables
-4x^2+x=0
a = -4; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-4)·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-4}=\frac{-2}{-8} =1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-4}=\frac{0}{-8} =0 $
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