10x^2+43x+28=

Solution for 10x^2+43x+28= equation:

10x^2+43x+28=

We move all terms to the left:

10x^2+43x+28-()=0

We add all the numbers together, and all the variables

10x^2+43x=0

a = 10; b = 43; c = 0;
Δ = b2-4ac
Δ = 432-4·10·0
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-43}{2*10}=\frac{-86}{20}
=-4+3/10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+43}{2*10}=\frac{0}{20}
=0
$