(2k-7)(k+5)=35

Solution for (2k-7)(k+5)=35 equation:

(2k-7)(k+5)=35

We move all terms to the left:

(2k-7)(k+5)-(35)=0

We multiply parentheses ..

(+2k^2+10k-7k-35)-35=0

We get rid of parentheses

2k^2+10k-7k-35-35=0

We add all the numbers together, and all the variables

2k^2+3k-70=0

a = 2; b = 3; c = -70;
Δ = b2-4ac
Δ = 32-4·2·(-70)
Δ = 569
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{569}}{2*2}=\frac{-3-\sqrt{569}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{569}}{2*2}=\frac{-3+\sqrt{569}}{4}
$