2x2-21x+45=

Solution for 2x2-21x+45= equation:

2x^2-21x+45=

We move all terms to the left:

2x^2-21x+45-()=0

We add all the numbers together, and all the variables

2x^2-21x=0

a = 2; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·2·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*2}=\frac{42}{4}
=10+1/2
$