4y^2-64y+255=0

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Solution for 4y^2-64y+255=0 equation:



4y^2-64y+255=0
a = 4; b = -64; c = +255;
Δ = b2-4ac
Δ = -642-4·4·255
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-4}{2*4}=\frac{60}{8} =7+1/2 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+4}{2*4}=\frac{68}{8} =8+1/2 $

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