10x^2+48x+54=0

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Solution for 10x^2+48x+54=0 equation: 



10x^2+48x+54=0

a = 10; b = 48; c = +54;
Δ = b2-4ac
Δ = 482-4·10·54
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12}{2*10}=\frac{-60}{20}
=-3
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12}{2*10}=\frac{-36}{20}
=-1+4/5
$

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