0.5(n+3)(n-4)=49

Solution for 0.5(n+3)(n-4)=49 equation:

0.5(n+3)(n-4)=49

We move all terms to the left:

0.5(n+3)(n-4)-(49)=0

We multiply parentheses ..

0.5(+n^2-4n+3n-12)-49=0

We multiply parentheses

0.5n^2-2n+1.5n-6-49=0

We add all the numbers together, and all the variables

0.5n^2-0.5n-55=0

a = 0.5; b = -0.5; c = -55;
Δ = b2-4ac
Δ = -0.52-4·0.5·(-55)
Δ = 110.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.5)-\sqrt{110.25}}{2*0.5}=\frac{0.5-\sqrt{110.25}}{1}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.5)+\sqrt{110.25}}{2*0.5}=\frac{0.5+\sqrt{110.25}}{1}
$