35x^2-42x+12=0

Solution for 35x^2-42x+12=0 equation:

35x^2-42x+12=0

a = 35; b = -42; c = +12;
Δ = b2-4ac
Δ = -422-4·35·12
Δ = 84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{21}}{2*35}=\frac{42-2\sqrt{21}}{70}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{21}}{2*35}=\frac{42+2\sqrt{21}}{70}
$