15=7t+(2t*2t)

Solution for 15=7t+(2t*2t) equation:

15=7t+(2t*2t)

We move all terms to the left:

15-(7t+(2t*2t))=0

We add all the numbers together, and all the variables

-(7t+(+2t*2t))+15=0


We calculate terms in parentheses: -(7t+(+2t*2t)), so:

7t+(+2t*2t)

We get rid of parentheses

7t+2t*2t

Wy multiply elements

4t^2+7t

Back to the equation:

-(4t^2+7t)


We get rid of parentheses

-4t^2-7t+15=0

a = -4; b = -7; c = +15;
Δ = b2-4ac
Δ = -72-4·(-4)·15
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*-4}=\frac{-10}{-8}
=1+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*-4}=\frac{24}{-8}
=-3
$