5((2^2x)+2x)=510

Solution for 5((2^2x)+2x)=510 equation:

5((2^2x)+2x)=510

We move all terms to the left:

5((2^2x)+2x)-(510)=0

We multiply parentheses

10x^2+10x-510=0

a = 10; b = 10; c = -510;
Δ = b2-4ac
Δ = 102-4·10·(-510)
Δ = 20500
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:

$\sqrt{\Delta}=\sqrt{20500}=\sqrt{100*205}=\sqrt{100}*\sqrt{205}=10\sqrt{205}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{205}}{2*10}=\frac{-10-10\sqrt{205}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{205}}{2*10}=\frac{-10+10\sqrt{205}}{20}
$