5h-293/2h+40=10

Solution for 5h-293/2h+40=10 equation:

5h-293/2h+40=10

We move all terms to the left:

5h-293/2h+40-(10)=0


Domain of the equation: 2h!=0

h!=0/2

h!=0

h∈R


We add all the numbers together, and all the variables

5h-293/2h+30=0

We multiply all the terms by the denominator


5h*2h+30*2h-293=0

Wy multiply elements

10h^2+60h-293=0

a = 10; b = 60; c = -293;
Δ = b2-4ac
Δ = 602-4·10·(-293)
Δ = 15320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{15320}=\sqrt{4*3830}=\sqrt{4}*\sqrt{3830}=2\sqrt{3830}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-2\sqrt{3830}}{2*10}=\frac{-60-2\sqrt{3830}}{20}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+2\sqrt{3830}}{2*10}=\frac{-60+2\sqrt{3830}}{20}
$