16x^2-40x+25=14x^2-47x+34

Solution for 16x^2-40x+25=14x^2-47x+34 equation:

16x^2-40x+25=14x^2-47x+34

We move all terms to the left:

16x^2-40x+25-(14x^2-47x+34)=0

We get rid of parentheses

16x^2-14x^2-40x+47x-34+25=0

We add all the numbers together, and all the variables

2x^2+7x-9=0

a = 2; b = 7; c = -9;
Δ = b2-4ac
Δ = 72-4·2·(-9)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-11}{2*2}=\frac{-18}{4}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+11}{2*2}=\frac{4}{4}
=1
$