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X+5.3=4X^2
We move all terms to the left:
X+5.3-(4X^2)=0
determiningTheFunctionDomain -4X^2+X+5.3=0
a = -4; b = 1; c = +5.3;
Δ = b2-4ac
Δ = 12-4·(-4)·5.3
Δ = 85.8
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{85.8}}{2*-4}=\frac{-1-\sqrt{85.8}}{-8} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{85.8}}{2*-4}=\frac{-1+\sqrt{85.8}}{-8} $
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