(3k-7)(k=-4)

Solution for (3k-7)(k=-4) equation:

(3k-7)(k=-4)

We move all terms to the left:

(3k-7)(k-(-4))=0


We calculate terms in parentheses: +(3k-7)(k-(-4)), so:

3k-7)(k-(-4)

We add all the numbers together, and all the variables

3k-7)(k+4

Back to the equation:

+(3k-7)(k+4)


We multiply parentheses ..

(+3k^2+12k-7k-28)=0

We get rid of parentheses

3k^2+12k-7k-28=0

We add all the numbers together, and all the variables

3k^2+5k-28=0

a = 3; b = 5; c = -28;
Δ = b2-4ac
Δ = 52-4·3·(-28)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-19}{2*3}=\frac{-24}{6}
=-4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+19}{2*3}=\frac{14}{6}
=2+1/3
$