1/4(10)^2t=5

Solution for 1/4(10)^2t=5 equation:

1/4(10)^2t=5

We move all terms to the left:

1/4(10)^2t-(5)=0


Domain of the equation: 410^2t!=0

t!=0/1

t!=0

t∈R


We multiply all the terms by the denominator


-5*410^2t+1=0

Wy multiply elements

-2050t^2+1=0

a = -2050; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-2050)·1
Δ = 8200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8200}=\sqrt{100*82}=\sqrt{100}*\sqrt{82}=10\sqrt{82}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{82}}{2*-2050}=\frac{0-10\sqrt{82}}{-4100}
=-\frac{10\sqrt{82}}{-4100}
=-\frac{\sqrt{82}}{-410}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{82}}{2*-2050}=\frac{0+10\sqrt{82}}{-4100}
=\frac{10\sqrt{82}}{-4100}
=\frac{\sqrt{82}}{-410}
$