Question

find the value of 'a' for the equation a^2+10a+2=26

Answer 1

Answer:

$a=-12,2$

Step-by-step explanation:

$a^{2} +10a-24=0\\(a+12)(a-2)=0\\a=-12, 2$

Answer 2

Answer:

a^2+10a+2=26

We move all terms to the left:

a^2+10a+2-(26)=0

We add all the numbers together, and all the variables

a^2+10a-24=0

a = 1; b = 10; c = -24;

Δ = b2-4ac

Δ = 102-4·1·(-24)

Δ = 196

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:

a1=−b−Δ√2aa2=−b+Δ√2a

Δ−−√=196−−−√=14

a1=−b−Δ√2a=−(10)−142∗1=−242=−12

a2=−b+Δ√2a=−(10)+142∗1=42=2