(-4k^2+14)=(k-7)

Solution for (-4k^2+14)=(k-7) equation:

(-4k^2+14)=(k-7)

We move all terms to the left:

(-4k^2+14)-((k-7))=0

We get rid of parentheses

-4k^2-((k-7))+14=0


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

(k-7)

We get rid of parentheses

k-7

Back to the equation:

-(k-7)


We get rid of parentheses

-4k^2-k+7+14=0

We add all the numbers together, and all the variables

-4k^2-1k+21=0

a = -4; b = -1; c = +21;
Δ = b2-4ac
Δ = -12-4·(-4)·21
Δ = 337
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{-(-1)-\sqrt{337}}{2*-4}=\frac{1-\sqrt{337}}{-8}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{337}}{2*-4}=\frac{1+\sqrt{337}}{-8}
$