4x^2-42x+76=0

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Solution for 4x^2-42x+76=0 equation:



4x^2-42x+76=0
a = 4; b = -42; c = +76;
Δ = b2-4ac
Δ = -422-4·4·76
Δ = 548
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:
$\sqrt{\Delta}=\sqrt{548}=\sqrt{4*137}=\sqrt{4}*\sqrt{137}=2\sqrt{137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{137}}{2*4}=\frac{42-2\sqrt{137}}{8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{137}}{2*4}=\frac{42+2\sqrt{137}}{8} $

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