2x^2-10x+3=0.

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Solution for 2x^2-10x+3=0. equation:



2x^2-10x+3=0.
a = 2; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·2·3
Δ = 76
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:
$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{19}}{2*2}=\frac{10-2\sqrt{19}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{19}}{2*2}=\frac{10+2\sqrt{19}}{4} $

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