Y=10x^2-140x-3

Solution for Y=10x^2-140x-3 equation:

=10Y^2-140Y-3

We move all terms to the left:

-(10Y^2-140Y-3)=0

We get rid of parentheses

-10Y^2+140Y+3=0

a = -10; b = 140; c = +3;
Δ = b2-4ac
Δ = 1402-4·(-10)·3
Δ = 19720
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{19720}=\sqrt{4*4930}=\sqrt{4}*\sqrt{4930}=2\sqrt{4930}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(140)-2\sqrt{4930}}{2*-10}=\frac{-140-2\sqrt{4930}}{-20}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(140)+2\sqrt{4930}}{2*-10}=\frac{-140+2\sqrt{4930}}{-20}
$