5(2x+1)+4=6(3x2)-7

Solution for 5(2x+1)+4=6(3x2)-7 equation:

5(2x+1)+4=6(3x^2)-7

We move all terms to the left:

5(2x+1)+4-(6(3x^2)-7)=0

We multiply parentheses

10x-(63x^2-7)+5+4=0

We get rid of parentheses

-63x^2+10x+7+5+4=0

We add all the numbers together, and all the variables

-63x^2+10x+16=0

a = -63; b = 10; c = +16;
Δ = b2-4ac
Δ = 102-4·(-63)·16
Δ = 4132
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:

$\sqrt{\Delta}=\sqrt{4132}=\sqrt{4*1033}=\sqrt{4}*\sqrt{1033}=2\sqrt{1033}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1033}}{2*-63}=\frac{-10-2\sqrt{1033}}{-126}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1033}}{2*-63}=\frac{-10+2\sqrt{1033}}{-126}
$