2x2+4x-48=0

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Solution for 2x2+4x-48=0 equation: 



2x^2+4x-48=0

a = 2; b = 4; c = -48;
Δ = b2-4ac
Δ = 42-4·2·(-48)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*2}=\frac{-24}{4}
=-6
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*2}=\frac{16}{4}
=4
$

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