15x^2=4(10x+3)

Solution for 15x^2=4(10x+3) equation:

15x^2=4(10x+3)

We move all terms to the left:

15x^2-(4(10x+3))=0


We calculate terms in parentheses: -(4(10x+3)), so:

4(10x+3)

We multiply parentheses

40x+12

Back to the equation:

-(40x+12)


We get rid of parentheses

15x^2-40x-12=0

a = 15; b = -40; c = -12;
Δ = b2-4ac
Δ = -402-4·15·(-12)
Δ = 2320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2320}=\sqrt{16*145}=\sqrt{16}*\sqrt{145}=4\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{145}}{2*15}=\frac{40-4\sqrt{145}}{30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{145}}{2*15}=\frac{40+4\sqrt{145}}{30}
$