2x^2-48x+190=0

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Solution for 2x^2-48x+190=0 equation: 



2x^2-48x+190=0

a = 2; b = -48; c = +190;
Δ = b2-4ac
Δ = -482-4·2·190
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-28}{2*2}=\frac{20}{4}
=5
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+28}{2*2}=\frac{76}{4}
=19
$

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