147-t^2=04

Solution for 147-t^2=04 equation:

147-t^2=04

We move all terms to the left:

147-t^2-(04)=0

We add all the numbers together, and all the variables

-1t^2+143=0

a = -1; b = 0; c = +143;
Δ = b2-4ac
Δ = 02-4·(-1)·143
Δ = 572
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{572}=\sqrt{4*143}=\sqrt{4}*\sqrt{143}=2\sqrt{143}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{143}}{2*-1}=\frac{0-2\sqrt{143}}{-2}
=-\frac{2\sqrt{143}}{-2}
=-\frac{\sqrt{143}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{143}}{2*-1}=\frac{0+2\sqrt{143}}{-2}
=\frac{2\sqrt{143}}{-2}
=\frac{\sqrt{143}}{-1}
$