(x^2+10)=(3x+17)

Solution for (x^2+10)=(3x+17) equation:

(x^2+10)=(3x+17)

We move all terms to the left:

(x^2+10)-((3x+17))=0

We get rid of parentheses

x^2-((3x+17))+10=0


We calculate terms in parentheses: -((3x+17)), so:

(3x+17)

We get rid of parentheses

3x+17

Back to the equation:

-(3x+17)


We get rid of parentheses

x^2-3x-17+10=0

We add all the numbers together, and all the variables

x^2-3x-7=0

a = 1; b = -3; c = -7;
Δ = b2-4ac
Δ = -32-4·1·(-7)
Δ = 37
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{37}}{2*1}=\frac{3-\sqrt{37}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{37}}{2*1}=\frac{3+\sqrt{37}}{2}
$