x^2-4=3x+24

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Solution for x^2-4=3x+24 equation:



x^2-4=3x+24
We move all terms to the left:
x^2-4-(3x+24)=0
We get rid of parentheses
x^2-3x-24-4=0
We add all the numbers together, and all the variables
x^2-3x-28=0
a = 1; b = -3; c = -28;
Δ = b2-4ac
Δ = -32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*1}=\frac{-8}{2} =-4 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*1}=\frac{14}{2} =7 $

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