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1=4x^2+3x
We move all terms to the left:
1-(4x^2+3x)=0
We get rid of parentheses
-4x^2-3x+1=0
a = -4; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·(-4)·1
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*-4}=\frac{-2}{-8} =1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*-4}=\frac{8}{-8} =-1 $
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