10x^2+103x+30=(x+10)

Solution for 10x^2+103x+30=(x+10) equation:

10x^2+103x+30=(x+10)

We move all terms to the left:

10x^2+103x+30-((x+10))=0


We calculate terms in parentheses: -((x+10)), so:

(x+10)

We get rid of parentheses

x+10

Back to the equation:

-(x+10)


We get rid of parentheses

10x^2+103x-x-10+30=0

We add all the numbers together, and all the variables

10x^2+102x+20=0

a = 10; b = 102; c = +20;
Δ = b2-4ac
Δ = 1022-4·10·20
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(102)-98}{2*10}=\frac{-200}{20}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(102)+98}{2*10}=\frac{-4}{20}
=-1/5
$