28g^2+27g=10

Solution for 28g^2+27g=10 equation:

28g^2+27g=10

We move all terms to the left:

28g^2+27g-(10)=0

a = 28; b = 27; c = -10;
Δ = b2-4ac
Δ = 272-4·28·(-10)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-43}{2*28}=\frac{-70}{56}
=-1+1/4
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+43}{2*28}=\frac{16}{56}
=2/7
$