(2x=5)(3x^2-10)

Solution for (2x=5)(3x^2-10) equation:

(2x=5)(3x^2-10)

We move all terms to the left:

(2x-(5)(3x^2-10))=0


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

2x-5(3x^2-10)

We multiply parentheses

-15x^2+2x+50

Back to the equation:

+(-15x^2+2x+50)


We get rid of parentheses

-15x^2+2x+50=0

a = -15; b = 2; c = +50;
Δ = b2-4ac
Δ = 22-4·(-15)·50
Δ = 3004
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:

$\sqrt{\Delta}=\sqrt{3004}=\sqrt{4*751}=\sqrt{4}*\sqrt{751}=2\sqrt{751}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{751}}{2*-15}=\frac{-2-2\sqrt{751}}{-30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{751}}{2*-15}=\frac{-2+2\sqrt{751}}{-30}
$