a^2-29.75a+49=0

Solution for a^2-29.75a+49=0 equation:

a^2-29.75a+49=0

a = 1; b = -29.75; c = +49;
Δ = b2-4ac
Δ = -29.752-4·1·49
Δ = 689.0625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29.75)-\sqrt{689.0625}}{2*1}=\frac{29.75-\sqrt{689.0625}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29.75)+\sqrt{689.0625}}{2*1}=\frac{29.75+\sqrt{689.0625}}{2}
$