2(2^x)+1/4(4^x)-320=0

Solution for 2(2^x)+1/4(4^x)-320=0 equation:

2(2^x)+1/4(4^x)-320=0


Domain of the equation: 44^x!=0

x!=0/1

x!=0

x∈R


We multiply all the terms by the denominator


22^x*44^x-320*44^x+1=0

Wy multiply elements

968x^2-14080x+1=0

a = 968; b = -14080; c = +1;
Δ = b2-4ac
Δ = -140802-4·968·1
Δ = 198242528
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:

$\sqrt{\Delta}=\sqrt{198242528}=\sqrt{1936*102398}=\sqrt{1936}*\sqrt{102398}=44\sqrt{102398}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14080)-44\sqrt{102398}}{2*968}=\frac{14080-44\sqrt{102398}}{1936}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14080)+44\sqrt{102398}}{2*968}=\frac{14080+44\sqrt{102398}}{1936}
$