49z^2-13=0

Solution for 49z^2-13=0 equation:

49z^2-13=0

a = 49; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·49·(-13)
Δ = 2548
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2548}=\sqrt{196*13}=\sqrt{196}*\sqrt{13}=14\sqrt{13}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{13}}{2*49}=\frac{0-14\sqrt{13}}{98}
=-\frac{14\sqrt{13}}{98}
=-\frac{\sqrt{13}}{7}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{13}}{2*49}=\frac{0+14\sqrt{13}}{98}
=\frac{14\sqrt{13}}{98}
=\frac{\sqrt{13}}{7}
$