(4n+5)(2n-7)=0

Solution for (4n+5)(2n-7)=0 equation:

(4n+5)(2n-7)=0

We multiply parentheses ..

(+8n^2-28n+10n-35)=0

We get rid of parentheses

8n^2-28n+10n-35=0

We add all the numbers together, and all the variables

8n^2-18n-35=0

a = 8; b = -18; c = -35;
Δ = b2-4ac
Δ = -182-4·8·(-35)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-38}{2*8}=\frac{-20}{16}
=-1+1/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+38}{2*8}=\frac{56}{16}
=3+1/2
$