49x^2-200x+16=0

Solution for 49x^2-200x+16=0 equation:

49x^2-200x+16=0

a = 49; b = -200; c = +16;
Δ = b2-4ac
Δ = -2002-4·49·16
Δ = 36864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36864}=192$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-200)-192}{2*49}=\frac{8}{98}
=4/49
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-200)+192}{2*49}=\frac{392}{98}
=4
$