28n2-47n+15=0

Solution for 28n2-47n+15=0 equation:

28n^2-47n+15=0

a = 28; b = -47; c = +15;
Δ = b2-4ac
Δ = -472-4·28·15
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-23}{2*28}=\frac{24}{56}
=3/7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+23}{2*28}=\frac{70}{56}
=1+1/4
$