6z^2+10z+5=99

Solution for 6z^2+10z+5=99 equation:

6z^2+10z+5=99

We move all terms to the left:

6z^2+10z+5-(99)=0

We add all the numbers together, and all the variables

6z^2+10z-94=0

a = 6; b = 10; c = -94;
Δ = b2-4ac
Δ = 102-4·6·(-94)
Δ = 2356
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:

$\sqrt{\Delta}=\sqrt{2356}=\sqrt{4*589}=\sqrt{4}*\sqrt{589}=2\sqrt{589}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{589}}{2*6}=\frac{-10-2\sqrt{589}}{12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{589}}{2*6}=\frac{-10+2\sqrt{589}}{12}
$