(3/4)b+(13/4)b=13

Solution for (3/4)b+(13/4)b=13 equation:

(3/4)b+(13/4)b=13

We move all terms to the left:

(3/4)b+(13/4)b-(13)=0


Domain of the equation: 4)b!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

(+3/4)b+(+13/4)b-13=0

We multiply parentheses

3b^2+13b^2-13=0

We add all the numbers together, and all the variables

16b^2-13=0

a = 16; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·16·(-13)
Δ = 832
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{13}}{2*16}=\frac{0-8\sqrt{13}}{32}
=-\frac{8\sqrt{13}}{32}
=-\frac{\sqrt{13}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{13}}{2*16}=\frac{0+8\sqrt{13}}{32}
=\frac{8\sqrt{13}}{32}
=\frac{\sqrt{13}}{4}
$