I=(-2x-1)(3x-5)

Solution for I=(-2x-1)(3x-5) equation:

=(-2I-1)(3I-5)

We move all terms to the left:

-((-2I-1)(3I-5))=0

We multiply parentheses ..

-((-6I^2+10I-3I+5))=0


We calculate terms in parentheses: -((-6I^2+10I-3I+5)), so:

(-6I^2+10I-3I+5)

We get rid of parentheses

-6I^2+10I-3I+5

We add all the numbers together, and all the variables

-6I^2+7I+5

Back to the equation:

-(-6I^2+7I+5)


We get rid of parentheses

6I^2-7I-5=0

a = 6; b = -7; c = -5;
Δ = b2-4ac
Δ = -72-4·6·(-5)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-13}{2*6}=\frac{-6}{12}
=-1/2
$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13}{2*6}=\frac{20}{12}
=1+2/3
$