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(z-15)(2z)=180
We move all terms to the left:
(z-15)(2z)-(180)=0
We multiply parentheses
2z^2-30z-180=0
a = 2; b = -30; c = -180;
Δ = b2-4ac
Δ = -302-4·2·(-180)
Δ = 2340
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2340}=\sqrt{36*65}=\sqrt{36}*\sqrt{65}=6\sqrt{65}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-6\sqrt{65}}{2*2}=\frac{30-6\sqrt{65}}{4} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+6\sqrt{65}}{2*2}=\frac{30+6\sqrt{65}}{4} $
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