4k^2+6k-15=3

Solution for 4k^2+6k-15=3 equation:

4k^2+6k-15=3

We move all terms to the left:

4k^2+6k-15-(3)=0

We add all the numbers together, and all the variables

4k^2+6k-18=0

a = 4; b = 6; c = -18;
Δ = b2-4ac
Δ = 62-4·4·(-18)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*4}=\frac{-24}{8}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*4}=\frac{12}{8}
=1+1/2
$