3a2+14a2=(3+14)=

Solution for 3a2+14a2=(3+14)= equation:

3a^2+14a^2=(3+14)=

We move all terms to the left:

3a^2+14a^2-((3+14))=0

We add all the numbers together, and all the variables

3a^2+14a^2-(17)=0

We add all the numbers together, and all the variables

17a^2-17=0

a = 17; b = 0; c = -17;
Δ = b2-4ac
Δ = 02-4·17·(-17)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-34}{2*17}=\frac{-34}{34}
=-1
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+34}{2*17}=\frac{34}{34}
=1
$