(3-z)(-4+z)=(2+z)z+7

Solution for (3-z)(-4+z)=(2+z)z+7 equation:

(3-z)(-4+z)=(2+z)z+7

We move all terms to the left:

(3-z)(-4+z)-((2+z)z+7)=0

We add all the numbers together, and all the variables

(-1z+3)(z-4)-((z+2)z+7)=0

We multiply parentheses ..

(-1z^2+4z+3z-12)-((z+2)z+7)=0


We calculate terms in parentheses: -((z+2)z+7), so:

(z+2)z+7

We multiply parentheses

z^2+2z+7

Back to the equation:

-(z^2+2z+7)


We get rid of parentheses

-1z^2-z^2+4z+3z-2z-12-7=0

We add all the numbers together, and all the variables

-2z^2+5z-19=0

a = -2; b = 5; c = -19;
Δ = b2-4ac
Δ = 52-4·(-2)·(-19)
Δ = -127
Delta is less than zero, so there is no solution for the equation