21x^2-71x=40

Solution for 21x^2-71x=40 equation:

21x^2-71x=40

We move all terms to the left:

21x^2-71x-(40)=0

a = 21; b = -71; c = -40;
Δ = b2-4ac
Δ = -712-4·21·(-40)
Δ = 8401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-71)-\sqrt{8401}}{2*21}=\frac{71-\sqrt{8401}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-71)+\sqrt{8401}}{2*21}=\frac{71+\sqrt{8401}}{42}
$