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4x^2+24x+19=3x^2
We move all terms to the left:
4x^2+24x+19-(3x^2)=0
determiningTheFunctionDomain 4x^2-3x^2+24x+19=0
We add all the numbers together, and all the variables
x^2+24x+19=0
a = 1; b = 24; c = +19;
Δ = b2-4ac
Δ = 242-4·1·19
Δ = 500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-10\sqrt{5}}{2*1}=\frac{-24-10\sqrt{5}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+10\sqrt{5}}{2*1}=\frac{-24+10\sqrt{5}}{2} $
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