47=5x+4x^2

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Solution for 47=5x+4x^2 equation:



47=5x+4x^2
We move all terms to the left:
47-(5x+4x^2)=0
We get rid of parentheses
-4x^2-5x+47=0
a = -4; b = -5; c = +47;
Δ = b2-4ac
Δ = -52-4·(-4)·47
Δ = 777
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{-(-5)-\sqrt{777}}{2*-4}=\frac{5-\sqrt{777}}{-8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{777}}{2*-4}=\frac{5+\sqrt{777}}{-8} $

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