25b^2+70b+48=0

Solution for 25b^2+70b+48=0 equation:

25b^2+70b+48=0

a = 25; b = 70; c = +48;
Δ = b2-4ac
Δ = 702-4·25·48
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-10}{2*25}=\frac{-80}{50}
=-1+3/5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+10}{2*25}=\frac{-60}{50}
=-1+1/5
$