49z2+40=41

Solution for 49z2+40=41 equation:

49z^2+40=41

We move all terms to the left:

49z^2+40-(41)=0

We add all the numbers together, and all the variables

49z^2-1=0

a = 49; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·49·(-1)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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